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; else pre->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
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 100 int 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