2014美团网笔试题目（总结）

前言

总结一下美团网笔试题目，明天可能去参加美团笔试

题目

1、一堆硬币，一个机器人，如果是反的就翻正，如果是正的就抛掷一次，无穷多次后，求正反的比例

解答：是不是题目不完整啊，我算的是3：1

2、一个汽车公司的产品，甲厂占40%，乙厂占60%，甲的次品率是1%，乙的次品率是2%，现在抽出一件汽车时次品，问是甲生产的可能性

解答：典型的贝叶斯公式，p(甲|废品) = p(甲 && 废品) / p(废品) = （0.4 × 0.01） /（0.4 × 0.01 + 0.6 × 0.02） = 0.25

3、k链表翻转。给出一个链表和一个数k，比如链表1→2→3→4→5→6，k=2，则翻转后2→1→4→3→6→5，若k=3,翻转后3→2→1→6→5→4，若k=4，翻转后4→3→2→1→5→6，用程序实现

解答：非递归可运行代码

#include <stdio.h>#include <stdlib.h>#include <string.h>typedef struct node {struct node *next;int data;} node;void createList(node **head, int data){node *pre, *cur, *new;pre = NULL;cur = *head;while (cur != NULL) {pre = cur;cur = cur->next;}new = (node *)malloc(sizeof(node));new->data = data;new->next = cur;if (pre == NULL)*head = new;elsepre->next = new;}void printLink(node *head){while (head->next != NULL) {printf("%d ", head->data);head = head->next;}printf("%d\n", head->data);}int linkLen(node *head){int len = 0;while (head != NULL) {len ++;head = head->next;}return len;}node* reverseK(node *head, int k){int i, len, time, now;len = linkLen(head);if (len < k) {return head;} else {time = len / k;}node *newhead, *prev, *next, *old, *tail;for (now = 0, tail = NULL; now < time; now ++) {old = head;for (i = 0, prev = NULL; i < k; i ++) {next = head->next;head->next = prev;prev = head;head = next;}if (now == 0) {newhead = prev;}old->next = head;if (tail != NULL) {tail->next = prev;}tail = old;}if (head != NULL) {tail->next = head;}return newhead;}int main(void){int i, n, k, data;node *head, *newhead;while (scanf("%d %d", &n, &k) != EOF) {for (i = 0, head = NULL; i < n; i ++) {scanf("%d", &data);createList(&head, data);}printLink(head);newhead = reverseK(head, k);printLink(newhead);}return 0;}

5、利用两个stack模拟queue

解答：剑指offer上的原题，九度oj有专门的练习，这里贴一下我的ac代码

#include <stdio.h>#include <stdlib.h>#include <string.h> typedef struct stack {    int top;    int seq[100000];} stack; /** * 入队操作 * * T = O(1) * */void pushQueue(stack *s1, int data){    s1->seq[s1->top ++] = data;} /** * 出队操作 * * T = O(n) * */void popQueue(stack *s1, stack *s2){    if (s2->top > 0) {        printf("%d\n", s2->seq[-- s2->top]);    } else {        while (s1->top > 0) {            s2->seq[s2->top ++] = s1->seq[-- s1->top];        }         if (s2->top > 0)             printf("%d\n", s2->seq[-- s2->top]);        else            printf("-1\n");    } }  int main(void){    int data, n;    stack *s1, *s2;    char str[5];     while (scanf("%d", &n) != EOF) {        // 初始化          s1 = (stack *)malloc(sizeof(stack));        s2 = (stack *)malloc(sizeof(stack));        s1->top = s2->top = 0;         while (n --) {            scanf("%s", str);             if (strcmp(str, "PUSH") == 0) { // 入队列                scanf("%d", &data);                pushQueue(s1, data);            } else {    // 出队列                popQueue(s1, s2);            }        }         free(s1);        free(s2);    }     return 0;}

6、一个m*n的矩阵，从左到右从上到下都是递增的，给一个数elem，求是否在矩阵中，给出思路和代码

解答：杨氏矩阵，简单题目 

#include <stdio.h>#include <stdlib.h>/** * 有序矩阵查找 * * T = O(n + n) * */void findKey(int **matrix, int n, int m, int key){int row, col;for (row = 0, col = m - 1; row < n && col >= 0;) {if (matrix[row][col] == key) {printf("第%d行，第%d列\n", row + 1, col + 1);break;} else if (matrix[row][col] > key) {col -= 1;} else {row += 1;}}printf("不存在!\n");}int main(void){int i, j, key, n, m, **matrix;// 构造矩阵scanf("%d %d", &n, &m);matrix = (int **)malloc(sizeof(int *) * n);for (i = 0; i < n; i ++)matrix[i] = (int *)malloc(sizeof(int) * m);for (i = 0; i < n; i ++) {for (j = 0; j < m; j ++)scanf("%d", &matrix[i][j]);}// 查询数据while (scanf("%d", &key) != EOF) {findKey(matrix, n, m, key);}return 0;}

7、编写函数，获取两段字符串的最长公共子串的长度，例如：
S1 = GCCCTAGCCAGDE
S2 = GCGCCAGTGDE
这两个序列的最长公共字串为GCCAG，也就是说返回值为5

解答：简单的动态规划题目，设dp[i][j]表示以s1[i]和s2[j]结尾的最长公共子串的长度，则状态转移方程为：

代码如下：

#include <stdio.h>#include <stdlib.h>#include <string.h>#define N 100int dp[N][N];void lcsLen(char *s1, char *s2, int len1, int len2){int i, j, max, index;memset(dp, 0, sizeof(dp));max = index = 0;for (i = 1; i <= len1; i ++) {for (j = 1; j <= len2; j ++) {if (s1[i] == s2[j]) {dp[i][j] = dp[i - 1][j - 1] + 1;if (dp[i][j] > max) {max = dp[i][j];index = i - max + 1;}} else {dp[i][j] = 0;}}}printf("最大长度为%d\n", max);for (i = 0; i < max; i ++) {printf("%c ", s1[i + index]);}printf("\n");}int main(void){char s1[N], s2[N];int i, len1, len2;while (scanf("%d %d", &len1, &len2) != EOF) {for (i = 1; i <= len1; i ++) {scanf("%c", &s1[i]);}for (i = 1; i <= len2; i ++) {scanf("%c", &s2[i]);}lcsLen(s1, s2, len1, len2);}return 0;}

8、有一个函数“int f(int n)”，请编写一段程序测试函数f(n)是否总是返回0，并添加必要的注释和说明

解答：博主对测试一向没有太大的兴趣，这道题让我考虑就是int从-2147483648-2147483647去遍历f的返回值，flag为标志位，不写代码了，太简单