• c语言经典算法大全 > 微软等数据结构 算法面试 100 题全部答案集锦
  • 微软等数据结构 算法面试 100 题全部答案集锦

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    微软等数据结构+ + + +算法面试100 100 100 100题全部答案集锦 作者:July、阿财. 时间:二零一一年十月十三日. 引言 无私分享造就开源的辉煌. 今是二零一一年十月十三日,明日14日即是本人刚好开博一周年.在一周年之际,特此分享出微软面试 全部100题答案的完整版,以作为对本博客所有读者的回馈. 一年之前的10月14日,一个名叫 July 的人在一个叫 csdn 的论坛上开帖分享微软等公司数据结构+算法 面试100题,自此,与上千网友一起做,一起思考,一起解答这些面试题目,最终成就了一个名为:结构之法 算法之道的编程面试与算法研究并重的博客,如今,此博客影响力逐步渗透到海外,及至到整个互联网. 在此之前,由于本人笨拙,这微软面试100题的答案只整理到了前60题(第1-60题答案可到本人资源下 载处下载:http://v_july_v.download.csdn.net/ http://v_july_v.download.csdn.net/ http://v_july_v.download.csdn.net/ http://v_july_v.download.csdn.net/),故此,常有朋友留言或来信询问后面40题的答案.只是 因个人认为:一、答案只是作为一个参考,不可太过依赖;二、常常因一些事情耽搁(如在整理最新的今年 九月、十月份的面试题:九月腾讯,创新工场,淘宝等公司最新面试十三题、十月百度,阿里巴巴,迅雷搜狗 最新面试十一题);三、个人正在针对那100题一题一题的写文章,多种思路,不断优化,即成程序员编程 艺术系列.自此,后面40题的答案迟迟未得整理.且个人已经整理的前60题的答案,在我看来,是有诸多问 题与弊端的,甚至很多答案都是错误的. 互联网总是能给人带来惊喜.前几日,一位现居美国加州的名叫阿财的朋友发来一封邮件,并把他自己 做的全部100题的答案一并发予给我,自此,便似遇见了知己.十分感谢. 任何东西只有分享出来才更显其价值.本只需贴出后面40题的答案,因为前60题的答案本人早已整理上 传至网上,但多一种思路多一种参考亦未尝不可.特此,把阿财的答案再稍加整理番,然后把全部100题的答 案现今都贴出来.若有任何问题,欢迎不吝指正.谢谢. 上千上万的人都关注过此100题,且大都都各自贡献了自己的思路,或回复于微软100 100 100 100题维护地址上,或 回复于本博客内,人数众多,无法一一标明,特此向他们诸位表示敬意和感谢.谢谢大家,诸君的努力足以影 响整个互联网,咱们已经迎来一个分享互利的新时代. 微软面试100 100 100 100题全部答案 最新整理的全部100题的答案参见如下(重复的,以及一些无关紧要的题目跳过.且因尊重阿财,未作过 多修改.因此,有些答案是有问题的,重点还可关注本人的程序员编程艺术系列,亦可参考个人之前 整理的前60题的答案:第1题-20题答案:http://blog.csdn.net/v_JULY_v/archive/2011/01/10/6126406.aspx, 第21-40题答案:http://blog.csdn.net/v_JULY_v/archive/2011/01/10/6126444.aspx,第41-60题答案: http://blog.csdn.net/v_JULY_v/archive/2011/02/01/6171539.aspx): 1.把二元查找树转变成排序的双向链表 题目: 输入一棵二元查找树,将该二元查找树转换成一个排序的双向链表. 要求不能创建任何新的结点,只调整指针的指向. 10 / \ 6 14 / \ / \ 4 8 12 16 转换成双向链表 4=6=8=10=12=14=16. 首先我们定义的二元查找树节点的数据结构如下: struct BSTreeNode { int m_nValue; // value of node BSTreeNode *m_pLeft; // left child of node BSTreeNode *m_pRight; // right child of node }; ANSWER: This is a traditional problem that can be solved using recursion. For each node, connect the double linked lists created from left and right child node to form a full list. /** * @param root The root node of the tree * @return The head node of the converted list. */ BSTreeNode * treeToLinkedList(BSTreeNode * root) { BSTreeNode * head, * tail; helper(head, tail, root); return head; } void helper(BSTreeNode *& head, BSTreeNode *& tail, BSTreeNode *root) { BSTreeNode *lt, *rh; if (root == NULL) { head = NULL, tail = NULL; return; } helper(head, lt, root->m_pLeft); helper(rh, tail, root->m_pRight); if (lt!=NULL) { lt->m_pRight = root; root->m_pLeft = lt; } else { head = root; } if (rh!=NULL) { root->m_pRight=rh; rh->m_pLeft = root; } else { tail = root; } } 2.设计包含 min 函数的栈. 定义栈的数据结构,要求添加一个 min 函数,能够得到栈的最小元素. 要求函数 min、push 以及 pop 的时间复杂度都是 O(1). ANSWER: Stack is a LIFO data structure. When some element is popped from the stack, the status will recover to the original status as before that element was pushed. So we can recover the minimum element, too. struct MinStackElement { int data; int min; }; struct MinStack { MinStackElement * data; int size; int top; } MinStack MinStackInit(int maxSize) { MinStack stack; stack.size = maxSize; stack.data = (MinStackElement*) malloc(sizeof(MinStackElement)*maxSize); stack.top = 0; return stack; } void MinStackFree(MinStack stack) { free(stack.data); } void MinStackPush(MinStack stack, int d) { if (stack.top == stack.size) error("out of stack space."); MinStackElement* p = stack.data[stack.top]; p->data = d; p->min = (stack.top==0?d : stack.data[top-1]); if (p->min > d) p->min = d; top ++; } int MinStackPop(MinStack stack) { if (stack.top == 0) error("stack is empty!"); return stack.data[--stack.top].data; } int MinStackMin(MinStack stack) { if (stack.top == 0) error("stack is empty!"); return stack.data[stack.top-1].min; } 3.求子数组的最大和 题目: 输入一个整形数组,数组里有正数也有负数. 数组中连续的一个或多个整数组成一个子数组,每个子数组都有一个和. 求所有子数组的和的最大值.要求时间复杂度为 O(n). 例如输入的数组为1, -2, 3, 10, -4, 7, 2, -5,和最大的子数组为3, 10, -4, 7, 2, 因此输出为该子数组的和18. ANSWER: A traditional greedy approach. Keep current sum, slide from left to right, when sum < 0, reset sum to 0. int maxSubarray(int a[], int size) { if (size<=0) error("error array size"); int sum = 0; int max = - (1 << 31); int cur = 0; while (cur < size) { sum += a[cur++]; if (sum > max) { max = sum; } else if (sum < 0) { sum = 0; } } return max; } 4.在二元树中找出和为某一值的所有路径 题目:输入一个整数和一棵二元树. 从树的根结点开始往下访问一直到叶结点所经过的所有结点形成一条路径. 打印出和与输入整数相等的所有路径. 例如输入整数22 和如下二元树 10 / \ 5 12 / \ 4 7 则打印出两条路径:10, 12 和10, 5, 7. 二元树节点的数据结构定义为: struct BinaryTreeNode // a node in the binary tree { int m_nValue; // value of node BinaryTreeNode *m_pLeft; // left child of node BinaryTreeNode *m_pRight; // right child of node }; ANSWER: Use backtracking and recurison. We need a stack to help backtracking the path. struct TreeNode { int data; TreeNode * left; TreeNode * right; }; void printPaths(TreeNode * root, int sum) { int path[MAX_HEIGHT]; helper(root, sum, path, 0); } void helper(TreeNode * root, int sum, int path[], int top) { path[top++] = root.data; sum -= root.data; if (root->left == NULL && root->right==NULL) { if (sum == 0) printPath(path, top); } else { if (root->left != NULL) helper(root->left, sum, path, top); if (root->right!=NULL) helper(root->right, sum, path, top); } top --; sum -= root.data; } 5.查找最小的 k 个元素 题目:输入 n 个整数,输出其中最小的 k 个. 例如输入1,2,3,4,5,6,7 和8 这8 个数字,则最小的4 个数字为1,2,3 和4. ANSWER: This is a very traditional question... O(nlogn): cat I_FILE | sort -n | head -n K O(kn): do insertion sort until k elements are retrieved. O(n+klogn): Take O(n) time to bottom-up build a min-heap. Then sift-down k-1 times. So traditional that I don't want to write the codes... Only gives the siftup and siftdown function. /** *@param i the index of the element in heap a[0...n-1] to be sifted up void siftup(int a[], int i, int n) { while (i>0) { int j=(i&1==0 ? i-1 : i+1); int p=(i-1)>>1; if (jnext; } while (h2->next != NULL) { h2 = h2-> next; } return h1 == h2; } // if there could exist cycle int isJoined(Node *h1, Node * h2) { Node* cylic1 = testCylic(h1); Node* cylic2 = testCylic(h2); if (cylic1+cylic2==0) return isJoinedSimple(h1, h2); if (cylic1==0 && cylic2!=0 || cylic1!=0 &&cylic2==0) return 0; Node *p = cylic1; while (1) { if (p==cylic2 || p->next == cylic2) return 1; p=p->next->next; cylic1 = cylic1->next; if (p==cylic1) return 0; } } Node* testCylic(Node * h1) { Node * p1 = h1, *p2 = h1; while (p2!=NULL && p2->next!=NULL) { p1 = p1->next; p2 = p2->next->next; if (p1 == p2) { return p1; } } return NULL; } 第8 题 此贴选一些比较怪的题,,由于其中题目本身与算法关系不大,仅考考思维.特此并作一题. 1.有两个房间,一间房里有三盏灯,另一间房有控制着三盏灯的三个开关, 这两个房间是分割开的,从一间里不能看到另一间的情况. 现在要求受训者分别进这两房间一次,然后判断出这三盏灯分别是由哪个开关控制的. 有什么办法呢? ANSWER: Skip. 2.你让一些人为你工作了七天,你要用一根金条作为报酬.金条被分成七小块,每天给出一 块. 如果你只能将金条切割两次,你怎样分给这些工人? ANSWER: 1+2+4; 3. 用一种算法来颠倒一个链接表的顺序.现在在不用递归式的情况下做一遍. ANSWER: Node * reverse(Node * head) { if (head == NULL) return head; if (head->next == NULL) return head; Node * ph = reverse(head->next); head->next->next = head; head->next = NULL; return ph; } Node * reverseNonrecurisve(Node * head) { if (head == NULL) return head; Node * p = head; Node * previous = NULL; while (p->next != NULL) { p->next = previous; previous = p; p = p->next; } p->next = previous; return p; } 用一种算法在一个循环的链接表里插入一个节点,但不得穿越链接表. ANSWER: I don't understand what is "Chuanyue". 用一种算法整理一个数组.你为什么选择这种方法? ANSWER: What is "Zhengli?" 用一种算法使通用字符串相匹配. ANSWER: What is "Tongyongzifuchuan"... a string with "*" and "?"? If so, here is the code. int match(char * str, char * ptn) { if (*ptn == '\0') return 1; if (*ptn do { if (match(str++, ptn+1)) return 1; } while (*str != '\0'); return 0; } if (*str == '\0') return 0; if (*str == *ptn || *ptn return match(str+1, ptn+1); } return 0; } 颠倒一个字符串.优化速度.优化空间. void reverse(char *str) { reverseFixlen(str, strlen(str)); } void reverseFixlen(char *str, int n) { char* p = str+n-1; while (str < p) { char c = *str; *str = *p; *p=c; } } 颠倒一个句子中的词的顺序,比如将"我叫克丽丝"转换为"克丽丝叫我", 实现速度最快,移动最少. ANSWER: Reverse the whole string, then reverse each word. Using the reverseFixlen() above. void reverseWordsInSentence(char * sen) { int len = strlen(sen); reverseFixlen(sen, len); char * p = str; while (*p!='\0') { while (*p p!='\0') p++; str = p; while (p!p!='\0') p++; reverseFixlen(str, p-str); } } 找到一个子字符串.优化速度.优化空间. ANSWER: KMP? BM? Sunday? Using BM or sunday, if it's ASCII string, then it's easy to fast access the auxiliary array. Otherwise an hashmap or bst may be needed. Lets assume it's an ASCII string. int bm_strstr(char *str, char *sub) { int len = strlen(sub); int i; int aux[256]; memset(aux, sizeof(int), 256, len+1); for (i=0; i=0 && str[j--] == sub[k--]) ; if (k<0) return j+1; if (i+1=len && strncmp(sub, p-len+1, len) == 0) return i-len; p++; } return -1; } 比较两个字符串,用O(n)时间和恒量空间. ANSWER: What is "comparing two strings"? Just normal string comparison? The natural way use O(n) time and O(1) space. int strcmp(char * p1, char * p2) { while (*p1 != '\0' && *p2 != '\0' && *p1 == *p2) { p1++, p2++; } if (*p1 == '\0' && *p2 == '\0') return 0; if (*p1 == '\0') return -1; if (*p2 == '\0') return 1; return (*p1 - *p2); // it can be negotiated whether the above 3 if's are necessary, I don't like to omit them. } 假设你有一个用1001 个整数组成的数组,这些整数是任意排列的,但是你知道所有的整数都在1 到1000(包括1000)之间.此外,除一个数字出现两次外,其他所有数字只出现一次.假设你只能对这个数组做 一次处理,用一种算法找出重复的那个数字.如果你在运算中使用了辅助的存储方式,那么你能找到不用这种 方式的算法吗? ANSWER: Sum up all the numbers, then subtract the sum from 1001*1002/2. Another way, use A XOR A XOR B = B: int findX(int a[]) { int k = a[0]; for (int i=1; i<=1000;i++) k ~= a[i]~i; } return k; } 不用乘法或加法增加8 倍.现在用同样的方法增加7 倍. ANSWER: n<<3; (n<<3)-n; 第9 题 判断整数序列是不是二元查找树的后序遍历结果 题目:输入一个整数数组,判断该数组是不是某二元查找树的后序遍历的结果. 如果是返回 true,否则返回 false. 例如输入5、7、6、9、11、10、8,由于这一整数序列是如下树的后序遍历结果: 8 / \ 6 10 / \ / \ 5 7 9 11 因此返回 true. 如果输入7、4、6、5,没有哪棵树的后序遍历的结果是这个序列,因此返回 false. ANSWER: This is an interesting one. There is a traditional question that requires the binary tree to be re-constructed from mid/post/pre order results. This seems similar. For the problems related to (binary) trees, recursion is the first choice. In this problem, we know in post-order results, the last number should be the root. So we have known the root of the BST is 8 in the example. So we can split the array by the root. int isPostorderResult(int a[], int n) { return helper(a, 0, n-1); } int helper(int a[], int s, int e) { if (e==s) return 1; int i=e-1; while (a[e]>a[i] && i>=s) i--; if (!helper(a, i+1, e-1)) return 0; int k = l; while (a[e]=s) i--; return helper(a, s, l); } 第10 题 翻转句子中单词的顺序. 题目:输入一个英文句子,翻转句子中单词的顺序,但单词内字符的顺序不变. 句子中单词以空格符隔开.为简单起见,标点符号和普通字母一样处理. 例如输入"I am a student.",则输出"student. a am I". Answer: Already done this. Skipped. 第11 题 求二叉树中节点的最大距离... 如果我们把二叉树看成一个图,父子节点之间的连线看成是双向的, 我们姑且定义"距离"为两节点之间边的个数. 写一个程序, 求一棵二叉树中相距最远的两个节点之间的距离. ANSWER: This is interesting... Also recursively, the longest distance between two nodes must be either from root to one leaf, or between two leafs. For the former case, it's the tree height. For the latter case, it should be the sum of the heights of left and right subtrees of the two leaves' most least ancestor. The first case is also the sum the heights of subtrees, just the height + 0. int maxDistance(Node * root) { int depth; return helper(root, depth); } int helper(Node * root, int &depth) { if (root == NULL) { depth = 0; return 0; } int ld, rd; int maxleft = helper(root->left, ld); int maxright = helper(root->right, rd); depth = max(ld, rd)+1; return max(maxleft, max(maxright, ld+rd)); } 第12 题 题目:求1+2+…+n, 要求不能使用乘除法、for、while、if、else、switch、case 等关键字以及条件判断语句 (A?B:C). ANSWER: 1+..+n=n*(n+1)/2=(n^2+n)/2 it is easy to get x/2, so the problem is to get n^2 though no if/else is allowed, we can easilly go around using short-pass. using macro to make it fancier: #define T(X, Y, i) (Y & (1<> 1; } 第13 题: 题目:输入一个单向链表,输出该链表中倒数第 k 个结点.链表的倒数第0 个结点为链表的尾指针. 链表结点定义如下: struct ListNode { int m_nKey; ListNode* m_pNext; }; Answer: Two ways. 1: record the length of the linked list, then go n-k steps. 2: use two cursors. Time complexities are exactly the same. Node * lastK(Node * head, int k) { if (k<0) error("k < 0"); Node *p=head, *pk=head; for (;k>0;k--) { if (pk->next!=NULL) pk = pk->next; else return NULL; } while (pk->next!=NULL) { p=p->next, pk=pk->next; } return p; } 第14 题: 题目:输入一个已经按升序排序过的数组和一个数字, 在数组中查找两个数,使得它们的和正好是输入的那个数字. 要求时间复杂度是 O(n).如果有多对数字的和等于输入的数字,输出任意一对即可. 例如输入数组1、2、4、7、11、15 和数字15.由于4+11=15,因此输出4 和11. ANSWER: Use two cursors. One at front and the other at the end. Keep track of the sum by moving the cursors. void find2Number(int a[], int n, int dest) { int *f = a, *e=a+n-1; int sum = *f + *e; while (sum != dest && f < e) { if (sum < dest) sum = *(++f); else sum = *(--e); } if (sum == dest) printf("%d, %d\n", *f, *e); } 第15 题: 题目:输入一颗二元查找树,将该树转换为它的镜像, 即在转换后的二元查找树中,左子树的结点都大于右子树的结点. 用递归和循环两种方法完成树的镜像转换. 例如输入: 8 / \ 6 10 /\ /\ 5 7 9 11 输出: 8 / \ 10 6 /\ /\ 11 9 7 5 定义二元查找树的结点为: struct BSTreeNode // a node in the binary search tree (BST) { int m_nValue; // value of node BSTreeNode *m_pLeft; // left child of node BSTreeNode *m_pRight; // right child of node }; ANSWER: This is the basic application of recursion. PS: I don't like the m_xx naming convension. void swap(Node ** l, Node ** r) { Node * p = *l; *l = *r; *r = p; } void mirror(Node * root) { if (root == NULL) return; swap(&(root->left), &(root->right)); mirror(root->left); mirror(root->right); } void mirrorIteratively(Node * root) { if (root == NULL) return; stack buf; buf.push(root); while (!stack.empty()) { Node * n = stack.pop(); swap(&(root->left), &(root->right)); if (root->left != NULL) buf.push(root->left); if (root->right != NULL) buf.push(root->right); } } 第16 题: 题目(微软): 输入一颗二元树,从上往下按层打印树的每个结点,同一层中按照从左往右的顺序打印. 例如输入 7 8 / \ 6 10 / \ / \ 5 7 9 11 输出8 6 10 5 7 9 11. ANSWER: The nodes in the levels are printed in the similar manner their parents were printed. So it should be an FIFO queue to hold the level. I really don't remember the function name of the stl queue, so I will write it in Java... void printByLevel(Node root) { Node sentinel = new Node(); LinkedList q=new LinkedList(); q.addFirst(root); q.addFirst(sentinel); while (!q.isEmpty()) { Node n = q.removeLast(); if (n==sentinel) { System.out.println("\n"); q.addFirst(sentinel); } else { System.out.println(n); if (n.left() != null) q.addFirst(n.left()); if (n.right()!=null) q.addFirst(n.right()); } } } 第17 题: 题目:在一个字符串中找到第一个只出现一次的字符.如输入 abaccdeff,则输出 b. 分析:这道题是2006 年google 的一道笔试题. ANSWER: Again, this depends on what is "char". Let's assume it as ASCII. char firstSingle(char * str) { int a[255]; memset(a, 0, 255*sizeof(int)); char *p=str; while (*p!='\0') { a[*p] ++; p++; } p = str; while (*p!='\0') { if (a[*p] == 1) return *p; } return '\0'; // this must the one that occurs exact 1 time. } 第18 题: 题目:n 个数字(0,1,…,n-1)形成一个圆圈,从数字0 开始, 每次从这个圆圈中删除第 m 个数字(第一个为当前数字本身,第二个为当前数字的下一个数 字). 当一个数字删除后,从被删除数字的下一个继续删除第 m 个数字. 求出在这个圆圈中剩下的最后一个数字. July:我想,这个题目,不少人已经见识过了. ANSWER: Actually, although this is a so traditional problem, I was always to lazy to think about this or even to search for the answer.(What a shame...). Finally, by google I found the elegant solution for it. The keys are: 1) if we shift the ids by k, namely, start from k instead of 0, we should add the result by k%n 2) after the first round, we start from k+1 ( possibly % n) with n-1 elements, that is equal to an (n-1) problem while start from (k+1)th element instead of 0, so the answer is (f(n-1, m)+k+1)%n 3) k = m-1, so f(n,m)=(f(n-1,m)+m)%n. finally, f(1, m) = 0; Now this is a O(n) solution. int joseph(int n, int m) { int fn=0; for (int i=2; i<=n; i++) { fn = (fn+m)%i; } return fn; } hu...长出一口气... 第19 题: 题目:定义 Fibonacci 数列如下: / 0 n=0 f(n)= 1 n=1 \ f(n-1)+f(n-2) n=2 输入 n,用最快的方法求该数列的第 n 项. 分析:在很多 C 语言教科书中讲到递归函数的时候,都会用 Fibonacci 作为例子. 因此很多程序员对这道题的递归解法非常熟悉,但....呵呵,你知道的.. ANSWER: This is the traditional problem of application of mathematics... let A= {1 1} {1 0} f(n) = A^(n-1)[0,0] this gives a O(log n) solution. int f(int n) { int A[4] = {1,1,1,0}; int result[4]; power(A, n, result); return result[0]; } void multiply(int[] A, int[] B, int _r) { _r[0] = A[0]*B[0] + A[1]*B[2]; _r[1] = A[0]*B[1] + A[1]*B[3]; _r[2] = A[2]*B[0] + A[3]*B[2]; _r[3] = A[2]*B[1] + A[3]*B[3]; } void power(int[] A, int n, int _r) { if (n==1) { memcpy(A, _r, 4*sizeof(int)); return; } int tmp[4]; power(A, n>>1, _r); multiply(_r, _r, tmp); if (n & 1 == 1) { multiply(tmp, A, _r); } else { memcpy(_r, tmp, 4*sizeof(int)); } } 第20 题: 题目:输入一个表示整数的字符串,把该字符串转换成整数并输出. 例如输入字符串"345",则输出整数345. ANSWER: This question checks how the interviewee is familiar with C/C++? I'm so bad at C/C++... int atoi(char * str) { int neg = 0; char * p = str; if (*p p++; neg = 1; } else if (*p p++; } int num = 0; while (*p != '\0') { if (*p>=0 && *p <= 9) { num = num * 10 + (*p-'0'); } else { error("illegal number"); } p++; } return num; } PS: I didn't figure out how to tell a overflow problem easily. 第21 题2010 年中兴面试题 编程求解: 输入两个整数 n 和m,从数列1,2,3.......n 中随意取几个数, 使其和等于 m ,要求将其中所有的可能组合列出来. ANSWER This is a combination generation problem. void findCombination(int n, int m) { if (n>m) findCombination(m, m); int aux[n]; memset(aux, 0, n*sizeof(int)); helper(m, 0, aux); } void helper(int dest, int idx, int aux[], int n) { if (dest == 0) dump(aux, n); if (dest <= 0 || idx==n) return; helper(dest, idx+1, aux, n); aux[idx] = 1; helper(dest-idx-1, idx+1, aux, n); aux[idx] = 0; } void dump(int aux[], int n) { for (int i=0; idata>h2->data) { head = h2; h2=h2->next; } else { head = h1; h1=h1->next; } Node * current = head; while (h1 != NULL && h2 != NULL) { if (h1 == NULL || (h2!=NULL && h1->data>h2->data)) { current->next = h2; h2=h2->next; current = current->next; } else { current->next = h1; h1=h1->next; current = current->next; } } current->next = NULL; return head; } 第25 题: 写一个函数,它的原形是 int continumax(char *outputstr,char *intputstr) 功能: 在字符串中找出连续最长的数字串,并把这个串的长度返回, 并把这个最长数字串付给其中一个函数参数 outputstr 所指内存. 例如:"abcd12345ed125ss123456789"的首地址传给 intputstr 后,函数将返回9, outputstr 所指的值为123456789 ANSWER: int continumax(char *outputstr, char *inputstr) { int len = 0; char * pstart = NULL; int max = 0; while (1) { if (*inputstr >= '0' && *inputstr <='9') { len ++; } else { if (len > max) pstart = inputstr-len; len = 0; } if (*inputstr++=='\0') break; } for (int i=0; i>= 8; } return c; } 29.栈的 push、pop 序列 题目:输入两个整数序列.其中一个序列表示栈的 push 顺序, 判断另一个序列有没有可能是对应的 pop 顺序. 为了简单起见,我们假设 push 序列的任意两个整数都是不相等的. 比如输入的 push 序列是1、2、3、4、5,那么4、5、3、2、1 就有可能是一个 pop 系列. 因为可以有如下的 push 和pop 序列: push 1,push 2,push 3,push 4,pop,push 5,pop,pop,pop,pop, 这样得到的 pop 序列就是4、5、3、2、1. 但序列4、3、5、1、2 就不可能是 push 序列1、2、3、4、5 的pop 序列. ANSWER This seems interesting. However, a quite straightforward and promising way is to actually build the stack and check whether the pop action can be achieved. int isPopSeries(int push[], int pop[], int n) { stack helper; int i1=0, i2=0; while (i2 < n) { while (stack.empty() || stack.peek() != pop[i2]) if (i1, a similar problem splits an array to halves as even as possible. It is possible to take binary search, when SUM of the array is not too high. Else this is a quite time consuming brute force problem. I cannot figure out a reasonable solution. 33. 实现一个挺高级的字符匹配算法: 给一串很长字符串,要求找到符合要求的字符串,例如目的串:123 1******3***2 ,12*****3 这些都要找出来 其实就是类似一些和谐系统..... ANSWER Not a clear problem. Seems a bitset can do. 34. 实现一个队列. 队列的应用场景为: 一个生产者线程将 int 类型的数入列,一个消费者线程将 int 类型的数出列 ANSWER I don't know multithread programming at all.... 35. 求一个矩阵中最大的二维矩阵(元素和最大).如: 1 2 0 3 4 2 3 4 5 1 1 1 5 3 0 中最大的是: 4 5 5 3 要求:(1)写出算法;(2)分析时间复杂度;(3)用C写出关键代码 ANSWER This is the traditional problem in Programming Pearls. However, the best result is too complicated to achieve. So lets do the suboptimal one. O(n^3) solution. 1) We have know that the similar problem for 1 dim array can be done in O(n) time. However, this cannot be done in both directions in the same time. We can only calculate the accumulations for all the sublist from i to j, (0<=i<=j acc(i,j) void accumulate(int a[], int n, int acc[]) { int i=0; acc[i] = a[i]; for (i=1;i1) { int i,j; for (i=0,j=0; im, generate a random number R=rand(N) in [0, N), replace a[R] with new number if R falls in [0, m). 3.大量的 URL 字符串,如何从中去除重复的,优化时间空间复杂度 ANSWER 1. Use hash map if there is enough memory. 2. If there is no enough memory, use hash to put urls to bins, and do it until we can fit the bin into memory. 39. 网易有道笔试: (1). 求一个二叉树中任意两个节点间的最大距离, 两个节点的距离的定义是这两个节点间边的个数, 比如某个孩子节点和父节点间的距离是1,和相邻兄弟节点间的距离是2,优化时间空间复 杂度. ANSWER Have done this. (2). 求一个有向连通图的割点,割点的定义是,如果除去此节点和与其相关的边, 有向图不再连通,描述算法. ANSWER Do dfs, record low[i] as the lowest vertex that can be reached from i and i's successor nodes. For each edge i, if low[i] = i and i is not a leaf in dfs tree, then i is a cut point. The other case is the root of dfs, if root has two or more children ,it is a cut point. /** * g is defined as: g[i][] is the out edges, g[i][0] is the edge count, g[i][1...g[i][0]] are the other end points. */ int cnt = 0; int visited[MAX_NUM]; int lowest[MAX_NUM]; void getCutPoints(int *g[], int cuts[], int n) { memset(cuts, 0, sizeof(int)*n); memset(visited, 0, sizeof(int)*n); memset(lowest, 0, sizeof(int)*n); for (int i=0; i visit[g[s][i]]) { low = visit[g[s][i]]; } } } lowest[s] = low; if (s == root && out > 1) { cuts[s] = 1; } return low; } 40.百度研发笔试题 引用自:zp155334877 1)设计一个栈结构,满足一下条件:min,push,pop 操作的时间复杂度为 O(1). ANSWER Have done this. 2)一串首尾相连的珠子(m 个),有N种颜色(N<=10), 设计一个算法,取出其中一段,要求包含所有 N 中颜色,并使长度最短. 并分析时间复杂度与空间复杂度. ANSWER Use a sliding window and a counting array, plus a counter which monitors the num of zero slots in counting array. When there is still zero slot(s), advance the window head, until there is no zero slot. Then shrink the window until a slot comes zero. Then one candidate segment of (window_size + 1) is achieved. Repeat this. It is O(n) algorithm since each item is swallowed and left behind only once, and either operation is in constant time. int shortestFullcolor(int a[], int n, int m) { int c[m], ctr = m; int h=0, t=0; int min=n; while (1) { while (ctr > 0 && h=n) return min; while (1) { c[a[t]] --; if (c[a[t]] == 0) break; t++; } if (min > h-t) min = h-t; t++; ctr++; } } 3)设计一个系统处理词语搭配问题,比如说中国和人民可以搭配, 则中国人民人民中国都有效.要求: *系统每秒的查询数量可能上千次; *词语的数量级为10W; *每个词至多可以与1W 个词搭配 当用户输入中国人民的时候,要求返回与这个搭配词组相关的信息. ANSWER This problem can be solved in three steps: 1. identify the words 2. recognize the phrase 3. retrieve the information Solution of 1: The most trivial way to efficiently identify the words is hash table or BST. A balanced BST with 100 words is about 17 levels high. Considering that 100k is not a big number, hashing is enough. Solution of 2: Since the phrase in this problem consists of only 2 words, it is easy to split the words. There won't be a lot of candidates. To find a legal combination, we need the "matching" information. So for each word, we need some data structure to tell whether a word can co-occur with it. 100k is a bad number -- cannot fit into a 16bit digit. However, 10k*100k is not too big, so we can simply use array of sorted array to do this. 1G integers, or 4G bytes is not a big number, We can also use something like VInt to save a lot of space. To find an index in a 10k sorted array, 14 comparisons are enough. Above operation can be done in any reasonable work-station's memory very fast, which should be the result of execution of about a few thousands of simple statements. Solution of 3: The information could be to big to fit in the memory. So a B-tree may be adopted to index the contents. Caching techniques is also helpful. Considering there are at most 10^9 entries, a 3 or 4 level of B- tree is okay, so it will be at most 5 disk access. However, there are thousands of requests and we can only do hundreds of disk seeking per second. It could be necessary to dispatch the information to several workstations. 41.求固晶机的晶元查找程序 晶元盘由数目不详的大小一样的晶元组成,晶元并不一定全布满晶元盘, 照相机每次这能匹配一个晶元,如匹配过,则拾取该晶元, 若匹配不过,照相机则按测好的晶元间距移到下一个位置. 求遍历晶元盘的算法求思路. ANSWER Dont understand. 42.请修改 append 函数,利用这个函数实现: 两个非降序链表的并集,1->2->3 和2->3->5 并为1->2->3->5 另外只能输出结果,不能修改两个链表的数据. ANSWER I don't quite understand what it means by "not modifying linked list's data". If some nodes will be given up, it is weird for this requirement. Node * head(Node *h1, Node * h2) { if (h1==NULL) return h2; if (h2==NULL) return h1; Node * head; if (h1->data < h2->data) { head =h1; h1=h1->next; } else { head = h2; h2=h2->next; } Node * p = head; while (h1!=NULL || h2!=NULL) { Node * candi; if (h1!=NULL && h2 != NULL && h1->data < h2->data || h2==NULL) { candi = h1; h1=h1->next; } else { candi = h2; h2=h2->next; } } if (candi->data == p->data) delete(candi); else { p->next = candi; p=candi; } } return head; } 43.递归和非递归俩种方法实现二叉树的前序遍历. ANSWER void preorderRecursive(TreeNode * node) { if (node == NULL) return; visit(node); preorderRecursive(node->left); preorderRecursive(node->right); } For non-recursive traversals, a stack must be adopted to replace the implicit program stack in recursive programs. void preorderNonrecursive(TreeNode * node) { stack s; s.push(node); while (!s.empty()) { TreeNode * n = s.pop(); visit(n); if (n->right!=NULL) s.push(n->right); if (n->left!=NULL) s.push(n->left); } } void inorderNonrecursive(TreeNode * node) { stack s; TreeNode * current = node; while (!s.empty() || current != NULL) { if (current != NULL) { s.push(current); current = current->left; } else { current = s.pop(); visit(current); current = current->right; } } } Postorder nonrecursive traversal is the hardest one. However, a simple observation helps that the node first traversed is the node last visited. This recalls the feature of stack. So we could use a stack to store all the nodes then pop them out altogether. This is a very elegant solution, while takes O(n) space. Other very smart methods also work, but this is the one I like the most. void postorderNonrecursive(TreeNode * node) { // visiting occurs only when current has no right child or last visited is his right child stack sTraverse, sVisit; sTraverse.push(node); while (!sTraverse.empty()) { TreeNode * p = sTraverse.pop(); sVisit.push(p); if (p->left != NULL) sTraverse.push(p->left); if (p->right != NULL) sTraverse.push(p->right); } while (!sVisit.empty()) { visit(sVisit.pop); } } 44.腾讯面试题: 1.设计一个魔方(六面)的程序. ANSWER This is a problem to test OOP. The object MagicCube must have following features 1) holds current status 2) easily doing transform 3) judge whether the final status is achieved 4) to test, it can be initialized 5) output current status public class MagicCube { // 6 faces, 9 chips each face private byte chips[54]; static final int X = 0; static final int Y = 1; static final int Z = 1; void transform(int direction, int level) { switch direction: { X : { transformX(level); break; } Y : { transformY(level); break; } Z : { transformZ(level); break; } default: throw new RuntimeException("what direction?"); } void transformX(int level) { … } } } // really tired of making this... } 2.有一千万条短信,有重复,以文本文件的形式保存,一行一条,有重复. 请用5 分钟时间,找出重复出现最多的前10 条. ANSWER 10M msgs, each at most 140 chars, that's 1.4G, which can fit to memory. So use hash map to accumulate occurrence counts. Then use a heap to pick maximum 10. 3.收藏了1 万条 url,现在给你一条 url,如何找出相似的 url.(面试官不解释何为相似) ANSWER What a SB interviewer... The company name should be claimed and if I met such a interviewer, I will contest to HR. The purpose of interview is to see the ability of communication. This is kind of single side shutdown of information exchange. My first answer will be doing edit distance to the url and every candidate. Then it depends on what interviewer will react. Other options includes: fingerprints, tries... 45.雅虎: 1.对于一个整数矩阵,存在一种运算,对矩阵中任意元素加一时,需要其相邻(上下左右) 某一个元素也加一,现给出一正数矩阵,判断其是否能够由一个全零矩阵经过上述运算得到. ANSWER A assignment problem. Two ways to solve. 1: duplicate each cell to as many as its value, do Hungarian algorithm. Denote the sum of the matrix as M, the edge number is 2M, so the complexity is 2*M*M; 2: standard maximum flow. If the size of matrix is NxN, then the algorithm using Ford Fulkerson algorithm is M*N*N. too complex... I will do this when I have time... 2.一个整数数组,长度为 n,将其分为 m 份,使各份的和相等,求m的最大值 比如{3,2,4,3,6} 可以分成{3,2,4,3,6} m=1; {3,6}{2,4,3} m=2 {3,3}{2,4}{6} m=3 所以 m 的最大值为3 ANSWER Two restrictions on m, 1) 1 <= m <= n; 2) Sum(array) mod m = 0 NOTE: no hint that a[i]>0, so m could be larger than sum/max; So firstly prepare the candidates, then do a brute force search on possible m's. In the search , a DP is available, since if f(array, m) = OR_i( f(array-subset(i), m) ), where Sum(subset(i)) = m. int maxShares(int a[], int n) { int sum = 0; int i, m; for (i=0; i=2; m--) { if (sum mod m != 0) continue; int aux[n]; for (i=0; i=0 && j >= 0; i--) { if (a[i] == ds[j]) { j --; } else if (a[i] < ds[j]) { ds[j--] = a[i]; } } return Arrays.copyOfRange(ds, 0, dsl+1); } 48.微软: 一个数组是由一个递减数列左移若干位形成的,比如{4,3,2,1,6,5} 是由{6,5,4,3,2,1}左移两位形成的,在这种数组中查找某一个数. ANSWER: The key is that, from the middle point of the array, half of the array is sorted, and the other half is a half-size shifted sorted array. So this can also be done recursively like a binary search. int shiftedBinarySearch(int a[], int k) { return helper(a, k, 0, n-1); } int helper(int a[], int k, int s, int t) { if (s>t) return -1; int m = s + (t-s)/2; if (a[m] == k) return m; else if (a[s] >= k && k > a[m]) return helper(a, k, s, m-1); else return helper(a, k, m+1, e); } 49.一道看上去很吓人的算法面试题: 如何对 n 个数进行排序,要求时间复杂度 O(n),空间复杂度 O(1) ANSWER: So a comparison sort is not allowed. Counting sort's space complexity is O(n). More ideas must be exchanged to find more conditions, else this is a crap. 50.网易有道笔试: 1.求一个二叉树中任意两个节点间的最大距离,两个节点的距离的定义是这两个节点间边 的个数, 比如某个孩子节点和父节点间的距离是1,和相邻兄弟节点间的距离是2,优化时间空间复 杂度. ANSWER: Have done this before. 2.求一个有向连通图的割点,割点的定义是, 如果除去此节点和与其相关的边,有向图不再连通,描述算法. ANSWER: Have done this before. 51.和为 n 连续正数序列. 题目:输入一个正数 n,输出所有和为 n 连续正数序列. 例如输入15,由于1+2+3+4+5=4+5+6=7+8=15,所以输出3 个连续序列1-5、4-6 和7-8. 分析:这是网易的一道面试题. ANSWER: It seems that this can be solved by factorization. However, factorization of large n is impractical! Suppose n=i+(i+1)+...+(j-1)+j, then n = (i+j)(j-i+1)/2 = (j*j - i*i + i + j)/2 => j^2 + j + (i-i^2-2n) = 0 => j=sqrt(i^2-i+1/4+2n) - 1/2 We know 1 <= i < j <= n/2 + 1 So for each i in [1, n/2], do this arithmetic to check if there is a integer answer. int findConsecutiveSequence(int n) { count = 0; for (int i=1; i<=n/2; i++) { int sqroot = calcSqrt(4*i*i+8*n-4*i+1); if (sqroot == -1) continue; if ((sqroot & 1) == 1) { System.out.println(i+sqroot-1)/2)); count ++; } } return count; } Use binary search to calculate sqrt, or just use math functions. 52.二元树的深度. 题目:输入一棵二元树的根结点,求该树的深度. 从根结点到叶结点依次经过的结点(含根、叶结点)形成树的一条路径,最长路径的长度为 树的深度. 例如:输入二元树: 10 / \ 6 14 / / \ 4 12 16 输出该树的深度3. 二元树的结点定义如下: struct SBinaryTreeNode // a node of the binary tree { int m_nValue; // value of node SBinaryTreeNode *m_pLeft; // left child of node SBinaryTreeNode *m_pRight; // right child of node }; 分析:这道题本质上还是考查二元树的遍历. ANSWER: Have done this. 53.字符串的排列. 题目:输入一个字符串,打印出该字符串中字符的所有排列. 例如输入字符串 abc,则输出由字符 a、b、c 所能排列出来的所有字符串 abc、acb、bac、bca、cab 和cba. 分析:这是一道很好的考查对递归理解的编程题, 因此在过去一年中频繁出现在各大公司的面试、笔试题中. ANSWER: Full permutation generation. I will use another technique that swap two neighboring characters each time. It seems that all the characters are different. I need to think about how to do it when duplications is allowed. Maybe simple recursion is better for that. void generatePermutation(char s[], int n) { if (n>20) { error("are you crazy?"); } byte d[n]; int pos[n], dpos[n]; // pos[i], the position of i'th number, dpos[i] the number in s[i] is the dpos[i]'th smallest qsort(s); // I cannot remember the form of qsort in C... memset(d, -1, sizeof(byte)*n); for (int i=0; idpos[r]; i--) d[i] = -d[i]; } } int findFirstAvailable(char s[], byte d[], int pos[], int n) { for (int i=n-1; i>1; i--) { if (s[pos[i]] > s[pos[i]+d[pos[i]]]) return pos[i]; } return -1; } #define aswap(ARR, X, Y) {int t=ARR[X]; ARR[X]=ARR[y]; ARR[Y]=t;} void swap(char s[], int pos[], int dpos[], byte d[], int r, int s) { aswap(s, r, s); aswap(d, r, s); aswap(pos, dpos[r], dpos[s]); aswap(dpos, r, s); } Maybe full of bugs. Please refer to algorithm manual for explansion. Pros: Amotized O(1) time for each move. Only two characters change position for each move. Cons: as you can see, very complicated. Extra space needed. 54.调整数组顺序使奇数位于偶数前面. 题目:输入一个整数数组,调整数组中数字的顺序,使得所有奇数位于数组的前半部分, 所有偶数位于数组的后半部分.要求时间复杂度为 O(n). ANSWER: This problem makes me recall the process of partition in quick sort. void partition(int a[], int n) { int i=j=0; while (i < n && (a[i] & 1)==0) i++; if (i==n) return; swap(a, i++, j++); while (i 1 + a[(i-1)*l2+j-1]) ? max : 1+a[(i-1)*l2+j-1]; } } } return a[l1*l2-1]; } 57.用俩个栈实现队列. 题目:某队列的声明如下: template class CQueue { public: CQueue() {} ~CQueue() {} void appendTail(const T& node); // append a element to tail void deleteHead(); // remove a element from head private: Stack m_stack1; Stack m_stack2; }; 分析:从上面的类的声明中,我们发现在队列中有两个栈. 因此这道题实质上是要求我们用两个栈来实现一个队列. 相信大家对栈和队列的基本性质都非常了解了:栈是一种后入先出的数据容器, 因此对队列进行的插入和删除操作都是在栈顶上进行;队列是一种先入先出的数据容器, 我们总是把新元素插入到队列的尾部,而从队列的头部删除元素. ANSWER Traditional problem in CLRS. void appendTail(const T& node) { m_stack1.push(node); } T getHead() { if (!m_stack2.isEmpty()) { return m_stack2.pop(); } if (m_stack1.isEmpty()) error("delete from empty queue"); while (!m_stack1.isEmpty()) { m_stack2.push(m_stack1.pop()); } return m_stack2.pop(); } 58.从尾到头输出链表. 题目:输入一个链表的头结点,从尾到头反过来输出每个结点的值.链表结点定义如下: struct ListNode { int m_nKey; ListNode* m_pNext; }; 分析:这是一道很有意思的面试题. 该题以及它的变体经常出现在各大公司的面试、笔试题中. ANSWER Have answered this... 59.不能被继承的类. 题目:用C++设计一个不能被继承的类. 分析:这是 Adobe 公司2007 年校园招聘的最新笔试题. 这道题除了考察应聘者的 C++基本功底外,还能考察反应能力,是一道很好的题目. ANSWER: I don't know c++. Maybe it can be done by implement an empty private default constructor. 60.在O(1)时间内删除链表结点. 题目:给定链表的头指针和一个结点指针,在O(1)时间删除该结点.链表结点的定义如下: struct ListNode { int m_nKey; ListNode* m_pNext; }; 函数的声明如下: void DeleteNode(ListNode* pListHead, ListNode* pToBeDeleted); 分析:这是一道广为流传的 Google 面试题,能有效考察我们的编程基本功,还能考察我们 的反应速度, 更重要的是,还能考察我们对时间复杂度的理解. ANSWER: Copy the data from tobedeleted's next to tobedeleted. then delete tobedeleted. The special case is tobedelete is the tail, then we must iterate to find its predecessor. The amortized time complexity is O(1). 61.找出数组中两个只出现一次的数字 题目:一个整型数组里除了两个数字之外,其他的数字都出现了两次. 请写程序找出这两个只出现一次的数字.要求时间复杂度是 O(n),空间复杂度是 O(1). 分析:这是一道很新颖的关于位运算的面试题. ANSWER: XOR. 62.找出链表的第一个公共结点. 题目:两个单向链表,找出它们的第一个公共结点. 链表的结点定义为: struct ListNode { int m_nKey; ListNode* m_pNext; }; 分析:这是一道微软的面试题.微软非常喜欢与链表相关的题目, 因此在微软的面试题中,链表出现的概率相当高. ANSWER: Have done this. 63.在字符串中删除特定的字符. 题目:输入两个字符串,从第一字符串中删除第二个字符串中所有的字符.例如,输入"They are students." 和"aeiou", 则删除之后的第一个字符串变成"Thy r stdnts.". 分析:这是一道微软面试题.在微软的常见面试题中,与字符串相关的题目占了很大的一部 分,因为写程序操作字符串能很好的反映我们的编程基本功. ANSWER: Have done this? Use a byte array / character hash to record second string. then use two pointers to shrink the 1st string. 64. 寻找丑数. 题目:我们把只包含因子2、3 和5 的数称作丑数(Ugly Number).例如6、8 都是丑数, 但14 不是,因为它包含因子7.习惯上我们把1 当做是第一个丑数.求按从小到大的顺序的第1500 个丑数. 分析:这是一道在网络上广为流传的面试题,据说 google 曾经采用过这道题. ANSWER: TRADITIONAL. Use heap/priority queue. int no1500() { int heap[4500]; heap[0] = 2; heap[1] = 3; heap[2] = 5; int size = 3; for (int i=1; i<1500; i++) { int s = heap[0]; heap[0] = s*2; siftDown(heap, 0, size); heap[size] = s*3; siftUp(heap, size, size+1); heap[size+1] = s*5; siftUp(heap, size+1, size+2); size+=2; } } void siftDown(int heap[], int from, int size) { int c = from * 2 + 1; while (c < size) { if (c+1 0) { int p = (from - 1) / 2; if (heap[p] > heap[from]) swap(heap, p, from); from = p; } } 65.输出1 到最大的 N 位数 题目:输入数字 n,按顺序输出从1 最大的 n 位10 进制数.比如输入3,则输出1、2、3 一直到最大的3 位 数即999. 分析:这是一道很有意思的题目.看起来很简单,其实里面却有不少的玄机. ANSWER: So maybe n could exceed i32? I cannot tell where is the trick... Who will output 2*10^9 numbers... 66.颠倒栈. 题目:用递归颠倒一个栈.例如输入栈{1, 2, 3, 4, 5},1 在栈顶. 颠倒之后的栈为{5, 4, 3, 2, 1},5 处在栈顶. ANSWER: Interesting... void reverse(Stack stack) { if (stack.size() == 1) return; Object o = stack.pop(); reverse(stack); putToBottom(stack, o); } void putToBottom(Stack stack, Object o) { if (stack.isEmpty()) { stack.push(o); return; } Object o2 = stack.pop(); putToBottom(stack, o); stack.push(o2); } 67.俩个闲玩娱乐. 1.扑克牌的顺子 从扑克牌中随机抽5 张牌,判断是不是一个顺子,即这5 张牌是不是连续的.2-10 为数字本身,A 为1,J 为11,Q 为12,K 为13,而大小王可以看成任意数字. ANSWER: // make king = 0 boolean isStraight(int a[]) { Arrays.sort(a); if (a[0] > 0) return checkGaps(a, 0, 4, 0); if (a[0] == 0 && a[1] != 0) return checkGaps(a, 1, 4, 1); return checkGaps(a, 2, 4, 2); } boolean checkGaps(int []a, int s, int e, int allowGaps) { int i=s; while (i6, f(m, n)=0 where m第一卷里面深入讲了排列的生成.这些算法的理解需要一定的数学功底,也需要一 定的灵感,有兴趣最好看看. ANSWER: Have done this. 71.数值的整数次方. 题目:实现函数 double Power(double base, int exponent),求base 的exponent 次方. 不需要考虑溢出. 分析:这是一道看起来很简单的问题.可能有不少的人在看到题目后30 秒写出如下的代码: double Power(double base, int exponent) { double result = 1.0; for(int i = 1; i <= exponent; ++i) result *= base; return result; } ANSWER … double power(double base, int exp) { if (exp == 1) return base; double half = power(base, exp >> 1); return (((exp & 1) == 1) ? base : 1.0) * half * half; } 72. 题目:设计一个类,我们只能生成该类的一个实例. 分析:只能生成一个实例的类是实现了 Singleton 模式的类型. ANSWER I'm not good at multithread programming... But if we set a lazy initialization, the "if" condition could be interrupted thus multiple constructor could be called, so we must add synchronized to the if judgements, which is a loss of efficiency. Putting it to the static initialization will guarantee that the constructor only be executed once by the java class loader. public class Singleton { private static Singleton instance = new Singleton(); private synchronized Singleton() { } public Singleton getInstance() { return instance(); } } This may not be correct. I'm quite bad at this... 73.对策字符串的最大长度. 题目:输入一个字符串,输出该字符串中对称的子字符串的最大长度.比如输入字符串"google",由于该字符 串里最长的对称子字符串是"goog",因此输出4. 分析:可能很多人都写过判断一个字符串是不是对称的函数,这个题目可以看成是该函数的 加强版. ANSWER Build a suffix tree of x and inverse(x), the longest anagram is naturally found. Suffix tree can be built in O(n) time so this is a linear time solution. 74.数组中超过出现次数超过一半的数字 题目:数组中有一个数字出现的次数超过了数组长度的一半,找出这个数字. 分析:这是一道广为流传的面试题,包括百度、微软和 Google 在内的多家公司都 曾经采用过这个题目.要几十分钟的时间里很好地解答这道题, 除了较好的编程能力之外,还需要较快的反应和较强的逻辑思维能力. ANSWER Delete every two different digits. The last one that left is the one. int getMajor(int a[], int n) { int x, cnt=0; for (int i=0; im_pLeft, X, Y); TreeNode * right = getLCA(root->m_pRight, X, Y); if (left == NULL) return right; else if (right == NULL) return left; else return root; } 76.复杂链表的复制 题目:有一个复杂链表,其结点除了有一个 m_pNext 指针指向下一个结点外,还有一个 m_pSibling 指向链 表中的任一结点或者 NULL.其结点的 C++定义如下: struct ComplexNode { int m_nValue; ComplexNode* m_pNext; ComplexNode* m_pSibling; }; 下图是一个含有5 个结点的该类型复杂链表. 图中实线箭头表示 m_pNext 指针,虚线箭头表示 m_pSibling 指针.为简单起见,指向 NULL 的指针没有画 出.请完成函数 ComplexNode* Clone(ComplexNode* pHead),以复制一个复杂链表. 分析:在常见的数据结构上稍加变化,这是一种很新颖的面试题. 要在不到一个小时的时间里解决这种类型的题目,我们需要较快的反应能力, 对数据结构透彻的理解以及扎实的编程功底. ANSWER Have heard this before, never seriously thought it. The trick is like this: take use of the old pSibling, make it points to the new created cloned node, while make the new cloned node's pNext backup the old pSibling. ComplexNode * Clone(ComplexNode* pHead) { if (pHead == NULL) return NULL; preClone(pHead); inClone(pHead); return postClone(pHead); } void preClone(ComplexNode* pHead) { ComplexNode * p = new ComplexNode(); p->m_pNext = pHead->m_pSibling; pHead->m_pSibling = p; if (pHead->m_pNext != NULL) preClone(pHead->m_pNext); } void inClone(ComplexNode * pHead) { ComplexNode * pSib = pNew->m_pNext; if (pSib == NULL) { pNew->m_pSibling = NULL; } else { pNew->m_pSibling = pSib->m_pSibling; } if (pHead->m_pNext != NULL) inClone(pHead->m_pNext); } ComplexNode * postClone(ComplexNode * pHead) { ComplexNode * pNew = pHead->m_pSibling; ComplexNode * pSib = pNew->m_pNext; if (pHead->m_pNext != NULL) { pNew->m_pNext = pHead->m_pNext->m_pSibling; pHead->m_pSibling = pSib; postClone(pHead->m_pNext); } else { pNew->pNext = NULL; pHead->m_pSibling = NULL; } return pNew; } 77.关于链表问题的面试题目如下: 1.给定单链表,检测是否有环. 使用两个指针 p1,p2 从链表头开始遍历,p1 每次前进一步,p2 每次前进两步.如果 p2 到 达链表尾部,说明无环,否则 p1、p2 必然会在某个时刻相遇(p1==p2),从而检测到链表中有环. 2.给定两个单链表(head1, head2),检测两个链表是否有交点,如果有返回第一个交点.如果 head1==head2,那么显然相交,直接返回 head1.否则,分别从 head1,head2 开始遍历两个链表获得其长 度len1 与len2,假设 len1>=len2,那么指针 p1 由head1 开始向后移动 len1-len2 步,指针 p2=head2,下面p1、p2 每次向后前进一步并比较 p1p2 是否相等,如果相等即返回该结点,否则说明两个链表没有交点. 3.给定单链表(head),如果有环的话请返回从头结点进入环的第一个节点. 运用题一,我们可以检查链表中是否有环.如果有环,那么 p1p2 重合点 p 必然在环中.从p点断开环,方 法为:p1=p, p2=p->next, p->next=NULL.此时,原单链表可以看作两条单链表,一条从 head 开始,另一条 从p2 开始,于是运用题二的方法,我们找到它们的第一个交点即为所求. 4.只给定单链表中某个结点 p(并非最后一个结点,即p->next!=NULL)指针,删除该结点.办法很简单,首先 是放 p 中数据,然后将 p->next 的数据 copy 入p中,接下来删除 p->next 即可. 5.只给定单链表中某个结点 p(非空结点),在p前面插入一个结点.办法与前者类似,首先分配一个结点 q, 将q插入在 p 后,接下来将 p 中的数据 copy 入q中,然后再将要插入的数据记录在 p 中. 78.链表和数组的区别在哪里? 分析:主要在基本概念上的理解. 但是最好能考虑的全面一点,现在公司招人的竞争可能就在细节上产生,谁比较仔细,谁获胜的机会就大. ANSWER 1. Besides the common staff, linked list is more abstract and array is usually a basic real world object. When mentioning "linked list", it doesn't matter how it is implemented, that is, as long as it supports "get data" and "get next", it is a linked list. But almost all programming languages provides array as a basic data structure. 2. So array is more basic. You can implement a linked list in an array, but cannot in the other direction. 79. 1.编写实现链表排序的一种算法.说明为什么你会选择用这样的方法? ANSWER For linked list sorting, usually mergesort is the best choice. Pros: O(1) auxilary space, compared to array merge sort. No node creation, just pointer operations. Node * linkedListMergeSort(Node * pHead) { int len = getLen(pHead); return mergeSort(pHead, len); } Node * mergeSort(Node * p, int len) { if (len == 1) { p->next = NULL; return p; } Node * pmid = p; for (int i=0; inext; } Node * p1 = mergeSort(p, len/2); Node * p2 = mergeSort(pmid, len - len/2); return merge(p1, p2); } Node * merge(Node * p1, Node * p2) { Node * p = NULL, * ph = NULL; while (p1!=NULL && p2!=NULL) { if (p1->datadata) { if (ph == NULL) {ph = p = p1;} else { p->next = p1; p1 = p1->next; p = p->next;} } else { if (ph == NULL) {ph = p = p2;} else { p->next = p2; p2 = p2->next; p = p->next;} } } p->next = (p1==NULL) ? p2 : p1; return ph; } 2.编写实现数组排序的一种算法.说明为什么你会选择用这样的方法? ANSWER Actually, it depends on the data. If arbitrary data is given in the array, I would choose quick sort. It is asy to implement, fast. 3.请编写能直接实现 strstr()函数功能的代码. ANSWER Substring test? Have done this. 80.阿里巴巴一道笔试题 问题描述: 12 个高矮不同的人,排成两排,每排必须是从矮到高排列,而且第二排比对应的第一排的人 高,问排列方式有多少种? 这个笔试题,很YD,因为把某个递归关系隐藏得很深. ANSWER Must be 1 a b … … c d e … … c could be 2th to 7th ( has to be smaller than d, e... those 5 numbers), so f(12) = 6 f(10) = 6* 5 f(8) = 30 * 4f(6) = 120*3f(4) = 360*2f(2) = 720 81.第1 组百度面试题 1.一个 int 数组,里面数据无任何限制,要求求出所有这样的数 a[i],其左边的数都小于等于它,右边的数都 大于等于它.能否只用一个额外数组和少量其它空间实现. ANSWER Sort the array to another array, compare it with the original array, all a[i] = b[i] are answers. 2.一个文件,内含一千万行字符串,每个字符串在1K 以内,要求找出所有相反的串对,如abc 和cba. ANSWER So we have ~10G data. It is unlikely to put them all into main memory. Anyway, calculate the hash of each line in the first round, at the second round calculate the hash of the reverse of the line and remembers only the line number pairs that the hashes of the two directions collides. The last round only test those lines. 3.STL 的set 用什么实现的?为什么不用 hash? ANSWER I don't quite know. Only heard of that map in stl is implemented with red-black tree. One good thing over hash is that you don't need to re-hash when data size grows. 82.第2 组百度面试题 1.给出两个集合 A 和B,其中集合 A={name}, 集合 B={age、sex、scholarship、address、...}, 要求: 问题1、根据集合 A 中的 name 查询出集合 B 中对应的属性信息; 问题2、根据集合 B 中的属性信息(单个属性,如age<20 等),查询出集合 A 中对应的 name. ANSWER SQL? Not a good defined question. 2.给出一个文件,里面包含两个字段{url、size},即url 为网址,size 为对应网址访问的次数 要求: 问题1、利用 Linux Shell 命令或自己设计算法,查询出 url 字符串中包含"baidu"子字符串对应的 size 字段 值; 问题2、根据问题1 的查询结果,对其按照 size 由大到小的排列. (说明:url 数据量很大,100 亿级以上) ANSWER 1. shell: gawk ' /baidu/ { print $2 } ' FILE 2. shell: gawk ' /baidu/ {print $2}' FILE | sort -n -r 83.第3 组百度面试题 1.今年百度的一道题目 百度笔试:给定一个存放整数的数组,重新排列数组使得数组左边为奇数,右边为偶数. 要求:空间复杂度 O(1),时间复杂度为 O(n). ANSWER Have done this. 2.百度笔试题 用C语言实现函数 void * memmove(void *dest, const void *src, size_t n).memmove 函数的功能是拷贝 src 所指的内存内容前 n 个字节到 dest 所指的地址上. 分析: 由于可以把任何类型的指针赋给 void 类型的指针, 这个函数主要是实现各种数据类型的拷贝. ANSWER //To my memory, usually memcpy doesn't check overlap, memmove do void * memmove(void * dest, const void * src, size_t n) { if (dest==NULL || src == NULL) error("NULL pointers"); byte * psrc = (byte*)src; byte * pdest = (byte*)dest; int step = 1; if (dest < src + n) { psrc = (byte*)(src+n-1); pdest = (byte*)(dest+n-1); step = -1; } for (int i=0; i= n and C(M-1, 2) < n. Do M rand()'s and get a binary string of M length. Assign 1100...0 to 1, 1010...0 to 2, ... 3.有10 个文件,每个文件1G, 每个文件的每一行都存放的是用户的 query,每个文件的 query 都可能重复. 要求按照 query 的频度排序. ANSWER If there is no enough memory, do bucketing first. For each bucket calculate the frequency of each query and sort. Then combine all the frequencies with multiway mergesort. 85.又见字符串的问题 1.给出一个函数来复制两个字符串 A 和B.字符串 A 的后几个字节和字符串 B 的前几个字节重叠.分析: 记住,这种题目往往就是考你对边界的考虑情况. ANSWER Special case of memmove. 2.已知一个字符串,比如 asderwsde,寻找其中的一个子字符串比如 sde 的个数,如果没有返回0,有的话返 回子字符串的个数. ANSWER ANSWER int count_of_substr(const char* str, const char * sub) { int count = 0; char * p = str; int n = strlen(sub); while ( *p != '\0' ) { if (strncmp(p, sub, n) == 0) count ++; p++; } return count; } Also recursive way works. Possible optimizations like Sunday algorithm or Rabin-Karp algorithm will do. 86. 怎样编写一个程序,把一个有序整数数组放到二叉树中? 分析:本题考察二叉搜索树的建树方法,简单的递归结构.关于树的算法设计一定要联想到递归,因为树本身 就是递归的定义.而,学会把递归改称非递归也是一种必要的技术.毕竟,递归会造成栈溢出,关于系统底层 的程序中不到非不得以最好不要用.但是对某些数学问题,就一定要学会用递归去解决. ANSWER This is the first question I'm given in a google interview. Node * array2Tree(int[] array) { return helper(array, 0, n-1); } Node * helper(int[] array, int start, int end) { if (start > end) return NULL; int m = start + (end-start)/2; Node * root = new Node(array[m]); root->left = helper(array, start, m-1); root->right = helper(array, m+1, end); return root; } 87. 1.大整数数相乘的问题.(这是2002 年在一考研班上遇到的算法题) ANSWER Do overflow manually. final static long mask = (1 << 31) - 1; ArrayList multiply(ArrayList a, ArrayList b) { ArrayList result = new ArrayList(a.size()*b.size()+1); for (int i=0; i x, int a, int base, ArrayList result) { if (a == 0) return; long overflow = 0; int i; for (i=0; i> 31); } while (overflow != 0) { long tmp = result.get(base+i) + overflow; result.set(base+i, (int) (mask & tmp)); overflow = (tmp >> 31); } } 2.求最大连续递增数字串(如"ads3sl456789DF3456ld345AA"中的"456789") ANSWER Have done this. 3.实现 strstr 功能,即在父串中寻找子串首次出现的位置. (笔试中常让面试者实现标准库中的一些函数) ANSWER Have done this. 88.2005 年11 月金山笔试题.编码完成下面的处理函数. 函数将字符串中的字符'*'移到串的前部分,前面的非'*'字符后移,但不能改变非'*'字符的先后顺序,函数返回 串中字符'*'的数量.如原始串为:ab**cd**e*12,处理后为*****abcde12,函数并返回值为5.(要求使用尽量 少的时间和辅助空间) ANSWER It's like partition in quick sort. Just keep the non-* part stable. int partitionStar(char a[]) { int count = 0; int i = a.length-1, j=a.length-1; // i for the cursor, j for the first non-* char while (i >= 0) { if (a[i] swap(a, i--, j--); } else { i--; count ++; } } return count; } 89.神州数码、华为、东软笔试题 1.2005 年11 月15 日华为软件研发笔试题.实现一单链表的逆转. ANSWER Have done this. 2.编码实现字符串转整型的函数(实现函数 atoi 的功能),据说是神州数码笔试题.如将字符 串"+123"123, "-0123"-123, "123CS45"123, "123.45CS"123, "CS123.45"0 ANSWER int atoi(const char * a) { if (*a=='+') return atoi(a+1); else if (*a=='-') return - atoi(a+1); char *p = a; int c = 0; while (*p >= '0' && *p <= '9') { c = c*10 + (*p - '0'); } return c; } 3.快速排序(东软喜欢考类似的算法填空题,又如堆排序的算法等) ANSWER Standard solution. Skip. 4.删除字符串中的数字并压缩字符串.如字符串"abc123de4fg56"处理后变为"abcdefg".注意空间和效率. (下面的算法只需要一次遍历,不需要开辟新空间,时间复杂度为 O(N)) ANSWER Also partition, keep non-digit stable. char * partition(const char * str) { char * i = str; // i for cursor, j for the first digit char; char * j = str; while (*i != '\0') { if (*i > '9' || *i < '0') { *j++ = *i++; } else { *i++; } } *j = '\0'; return str; } 5.求两个串中的第一个最长子串(神州数码以前试题). 如"abractyeyt","dgdsaeactyey"的最大子串为"actyet". ANSWER Use suffix tree. The longest common substring is the longest prefix of the suffixes. O(n) to build suffix tree. O(n) to find the lcs. 90. 1.不开辟用于交换数据的临时空间,如何完成字符串的逆序 (在技术一轮面试中,有些面试官会这样问). ANSWER Two cursors. 2.删除串中指定的字符 (做此题时,千万不要开辟新空间,否则面试官可能认为你不适合做嵌入式开发) ANSWER Have done this. 3.判断单链表中是否存在环. ANSWER Have done this. 91 1.一道著名的毒酒问题 有1000 桶酒,其中1 桶有毒.而一旦吃了,毒性会在1 周后发作.现在我们用小老鼠做实验,要在1 周内找 出那桶毒酒,问最少需要多少老鼠. ANSWER Have done this. 10 mices. 2.有趣的石头问题 有一堆1 万个石头和1 万个木头,对于每个石头都有1 个木头和它重量一样, 把配对的石头和木头找出来. ANSWER Quick sort. 92. 1.多人排成一个队列,我们认为从低到高是正确的序列,但是总有部分人不遵守秩序.如果说,前面的人比后面的 人高(两人身高一样认为是合适的), 那么我们就认为这两个人是一对"捣乱分子",比如说,现在存在一个序列: 176, 178, 180, 170, 171 这些捣乱分子对为 <176, 170>, <176, 171>, <178, 170>, <178, 171>, <180, 170>, <180, 171>, 那么,现在给出一个整型序列,请找出这些捣乱分子对的个数(仅给出捣乱分子对的数目即可,不用具体的对) 要求: 输入: 为一个文件(in),文件的每一行为一个序列.序列全为数字,数字间用","分隔. 输出: 为一个文件(out),每行为一个数字,表示捣乱分子的对数. 详细说明自己的解题思路,说明自己实现的一些关键点. 并给出实现的代码,并分析时间复杂度. 限制: 输入每行的最大数字个数为100000 个,数字最长为6 位.程序无内存使用限制. ANSWER The answer is the swap number of insertion sort. The straightforward method is to do insertion sort and accumulate the swap numbers, which is slow: O(n^2) A sub-quadratic solution can be done by DP. f(n) = f(n-1) + Index(n) Index(n), which is to determine how many numbers is smaller than a[n] in a[0..n-1], can be done in log(n) time using BST with subtree size. 93.在一个 int 数组里查找这样的数,它大于等于左侧所有数,小于等于右侧所有数.直观想法是用两个数组 a、b.a[i]、b[i]分别保存从前到 i 的最大的数和从后到 i 的最小的数,一个解答:这需要两次遍历,然后再遍 历一次原数组,将所有 data[i]>=a[i-1]&&data[i]<=b[i]的data[i]找出即可.给出这个解答后,面试官有要求只能 用一个辅助数组,且要求少遍历一次. ANSWER It is natural to improve the hint... just during the second traversal, do the range minimum and picking together. There is no need to store the range minimums. 94.微软笔试题 求随机数构成的数组中找到长度大于=3 的最长的等差数列, 输出等差数列由小到大: 如果没有符合条件的就输出 格式: 输入[1,3,0,5,-1,6] 输出[-1,1,3,5] 要求时间复杂度,空间复杂度尽量小 ANSWER Firstly sort the array. Then do DP: for each a[i], update the length of the arithmetic sequences. That's a O(n^3) solution. Each arithmetic sequence can be determined by the last item and the step size. 95.华为面试题 1 判断一字符串是不是对称的,如:abccba ANSWER Two cursors. 2.用递归的方法判断整数组 a[N]是不是升序排列 ANSWER boolean isAscending(int a[]) { return isAscending(a, 0); } boolean isAscending(int a[], int start) { return start == a.length - 1 || isAscending(a, start+1); } 96.08 年中兴校园招聘笔试题 1.编写 strcpy 函数 已知 strcpy 函数的原型是 char *strcpy(char *strDest, const char *strSrc); 其中 strDest 是目的字符串,strSrc 是源字符串.不调用 C++/C 的字符串库函数,请编写函数 strcpy ANSWER char *strcpy(char *strDest, const char *strSrc) { if (strSrc == NULL) return NULL; char *i = strSrc, *j = strDest; while (*i != '\0') { *j++ = *i++; } *j = '\0'; return strDest; } Maybe you need to check if src and dest overlaps, then decide whether to copy from tail to head. 最后压轴之戏,终结此微软等100 题系列 V0.1 版. 那就, 连续来几组微软公司的面试题,让你一次爽个够: 97.第1 组微软较简单的算法面试题 1.编写反转字符串的程序,要求优化速度、优化空间. ANSWER Have done this. 2.在链表里如何发现循环链接? ANSWER Have done this. 3.编写反转字符串的程序,要求优化速度、优化空间. ANSWER Have done this. 4.给出洗牌的一个算法,并将洗好的牌存储在一个整形数组里. ANSWER Have done this. 5.写一个函数,检查字符是否是整数,如果是,返回其整数值. (或者:怎样只用4 行代码编写出一个从字符串到长整形的函数?) ANSWER Char or string? have done atoi; 98.第2 组微软面试题 1.给出一个函数来输出一个字符串的所有排列. ANSWER Have done this... 2.请编写实现 malloc()内存分配函数功能一样的代码. ANSWER Way too hard as an interview question... Please check wikipedia for solutions... 3.给出一个函数来复制两个字符串 A 和B.字符串 A 的后几个字节和字符串 B 的前几个字节重叠. ANSWER Copy from tail to head. 4.怎样编写一个程序,把一个有序整数数组放到二叉树中? ANSWER Have done this. 5.怎样从顶部开始逐层打印二叉树结点数据?请编程. ANSWER Have done this... 6.怎样把一个链表掉个顺序(也就是反序,注意链表的边界条件并考虑空链表)? ANSWER Have done this... 99.第3 组微软面试题 1.烧一根不均匀的绳,从头烧到尾总共需要1 个小时.现在有若干条材质相同的绳子,问如何用烧绳的方法来 计时一个小时十五分钟呢? ANSWER May have done this... burn from both side gives ? hour. 2.你有一桶果冻,其中有黄色、绿色、红色三种,闭上眼睛抓取同种颜色的两个.抓取多少个就可以确定你肯 定有两个同一颜色的果冻?(5 秒-1 分钟) ANSWER 4. 3.如果你有无穷多的水,一个3 公升的提捅,一个5 公升的提捅,两只提捅形状上下都不均 匀,问你如何才能准确称出4 公升的水?(40 秒-3 分钟) ANSWER 5 to 3 => 2 2 to 3, remaining 1 5 to remaining 1 => 4 一个岔路口分别通向诚实国和说谎国. 来了两个人,已知一个是诚实国的,另一个是说谎国的. 诚实国永远说实话,说谎国永远说谎话.现在你要去说谎国, 但不知道应该走哪条路,需要问这两个人.请问应该怎么问?(20 秒-2 分钟) ANSWER Seems there are too many answers. I will pick anyone to ask: how to get to your country? Then pick the other way. 100.第4 组微软面试题,挑战思维极限 1.12 个球一个天平,现知道只有一个和其它的重量不同,问怎样称才能用三次就找到那个 球.13 个呢?(注意此题并未说明那个球的重量是轻是重,所以需要仔细考虑)(5 分钟-1 小时) ANSWER Too complicated. Go find brain teaser answers by yourself. 2.在9 个点上画10 条直线,要求每条直线上至少有三个点?(3 分钟-20 分钟) 3.在一天的24 小时之中,时钟的时针、分针和秒针完全重合在一起的时候有几次?都分别是什么时间?你怎 样算出来的?(5 分钟-15 分钟) 30 终结附加题: 微软面试题,挑战你的智商 说明:如果你是第一次看到这种题,并且以前从来没有见过类似的题型, 并且能够在半个小时之内做出答案,说明你的智力超常..) 1.第一题. 五个海盗抢到了100 颗宝石,每一颗都一样大小和价值连城.他们决定这么分: 抽签决定自己的号码(1、2、3、4、5) 首先,由1 号提出分配方案,然后大家表决,当且仅当超过半数的人同意时, 按照他的方案进行分配,否则将被扔进大海喂鲨鱼 如果1 号死后,再由2 号提出分配方案,然后剩下的4 人进行表决, 当且仅当超过半数的人同意时,按照他的方案进行分配,否则将被扔入大海喂鲨鱼. 依此类推 条件:每个海盗都是很聪明的人,都能很理智地做出判断,从而做出选择. 问题:第一个海盗提出怎样的分配方案才能使自己的收益最大化? Answer: A traditional brain teaser. Consider #5, whatever #4 proposes, he won't agree, so #4 must agree whatever #3 proposes. So if there are only #3-5, #3 should propose (100, 0, 0). So the expected income of #3 is 100, and #4 and #5 is 0 for 3 guy problem. So whatever #2 proposes, #3 won't agree, but if #2 give #4 and #5 $1, they can get more than 3-guy subproblem. So #2 will propose (98, 0, 1, 1). So for #1, if give #2 less than $98, #2 won't agree. But he can give #3 $1 and #4 or #5 $2, so this is a (97, 0, 1, 2, 0) solution. 2.一道关于飞机加油的问题,已知: 每个飞机只有一个油箱, 飞机之间可以相互加油(注意是相互,没有加油机) 一箱油可供一架飞机绕地球飞半圈, 问题: 为使至少一架飞机绕地球一圈回到起飞时的飞机场,至少需要出动几架飞机? (所有飞机从同一机场起飞,而且必须安全返回机场,不允许中途降落,中间没有飞机场) Pass.ok,微软面试全部100题答案至此完. 后记 2010已过,如今个人早已在整理2011最新的面试题,参见如下: ? 微软、谷歌、百度等公司经典面试100题[第1-60题] (微软100题第二版前60题) ? 微软、Google 等公司非常好的面试题及解答[第61-70题] (微软100题第二版第61-70题) ? 十道海量数据处理面试题与十个方法大总结 (十道海量数据处理面试题) ? 海量数据处理面试题集锦与 Bit-map 详解 (十七道海量数据处理面试题) ? 九月腾讯,创新工场,淘宝等公司最新面试十三题(2011年度9月最新面试30题) ? 十月百度,阿里巴巴,迅雷搜狗最新面试十一题(2011年度十月最新面试题集锦) 一切的详情,可看此文:横空出世,席卷 Csdn-- Csdn-- Csdn-- Csdn--评微软等数据结构+ + + +算法面试100 100 100 100题 (在此文中,你能 找到与微软100题所有一切相关的东西).资源下载和维护地址分别如下所示: ? 所有的资源下载(题目+答案)地址:http://v_july_v.download.csdn.net/ http://v_july_v.download.csdn.net/ http://v_july_v.download.csdn.net/ http://v_july_v.download.csdn.net/. ? 本微软等100 题系列 V0.1 版,永久维护地址:http://topic.csdn.net/u/20101126/10/b4f12a00- http://topic.csdn.net/u/20101126/10/b4f12a00- http://topic.csdn.net/u/20101126/10/b4f12a00- http://topic.csdn.net/u/20101126/10/b4f12a00- 6280-492f-b785-cb6835a63dc9.html 6280-492f-b785-cb6835a63dc9.html 6280-492f-b785-cb6835a63dc9.html 6280-492f-b785-cb6835a63dc9.html. 欢迎,任何人,就以上任何内容,题目与答案思路,或其它任何问题、与我联系.本人邮箱: zhoulei0907@yahoo.cn. 作者声明: 本人 July 对以上所有任何内容和资料享有版权,转载请注明作者本人 July 及出处. 向你的厚道致敬.谢谢.二零一一年十月十三日、以诸君为傲.
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