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2014美團(tuán)網(wǎng)筆試題目
1、一堆硬幣，一個(gè)機(jī)器人，如果是反的就翻正，如果是正的就拋擲一次，無窮多次后，求正反的比例
解答：是不是題目不完整啊，我算的是3：1
2、一個(gè)汽車公司的產(chǎn)品，甲廠占40%，乙廠占60%，甲的次品率是1%，乙的次品率是2%，現(xiàn)在抽出一件汽車時(shí)次品，問是甲生產(chǎn)的可能性
解答：典型的貝葉斯公式，p(甲|廢品) = p(甲 && 廢品) / p(廢品) = (0.4 × 0.01) /(0.4 × 0.01 + 0.6 × 0.02) = 0.25
3、k鏈表翻轉(zhuǎn)。給出一個(gè)鏈表和一個(gè)數(shù)k，比如鏈表1→2→3→4→5→6，k=2，則翻轉(zhuǎn)后2→1→4→3→6→5，若k=3,翻轉(zhuǎn)后3→2→1→6→5→4，若k=4，翻轉(zhuǎn)后4→3→2→1→5→6，用程序?qū)崿F(xiàn)
非遞歸可運(yùn)行代碼：
#include
#include
#include
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、利用兩個(gè)stack模擬queue
劍指offer上的原題，九度oj有專門的練習(xí)，這里貼一下我的ac代碼：
#include
#include
#include
typedef struct stack {
int top;
int seq[100000];
} stack;
/**
* 入隊(duì)操作
*
* T = O(1)
*
*/
void pushQueue(stack *s1, int data)
{
s1->seq[s1->top ++] = data;
}
/**
* 出隊(duì)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) { // 入隊(duì)列
scanf("%d", &data);
pushQueue(s1, data);
} else { // 出隊(duì)列
popQueue(s1, s2);
}
}
free(s1);
free(s2);
}
return 0;
}
6、一個(gè)m*n的矩陣，從左到右從上到下都是遞增的，給一個(gè)數(shù)elem，求是否在矩陣中，給出思路和代碼
楊氏矩陣，簡(jiǎn)單題目：
#include
#include
/**
* 有序矩陣查找
*
* 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\(zhòng)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;
// 構(gòu)造矩陣
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]);
}
// 查詢數(shù)據(jù)
while (scanf("%d", &key) != EOF) {
findKey(matrix, n, m, key);
}
return 0;
}
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