c语言经典面试题

来源：互联网发布：南通大学网络教学平台编辑：程序博客网时间：2024/06/06 01:09

这里写图片描述

上面两幅图中有六道经典C语言面试题，真的可以认真看看！！！

直接贴代码：

#include<stdio.h>#include<windows.h>#include<assert.h>void bit_set(unsigned char *p_data, unsigned char position, int flag){    assert(p_data!=NULL);    if ((position >= 1) && (position <= 8))//指定位置1-8    {        if (0 == flag)        {            (*p_data) &= ~(1 << (position - 1));        }        else if (1 == flag)        {            (*p_data) |= (1 << (position - 1));        }    }}void test(){    unsigned char i = 15;    bit_set(&i, 3, 0);    printf("%d\n", i);}void  rightloopmove(char *str, unsigned short steps){    assert(str != NULL);    int i = 0;    int j = 0;    int len = strlen(str) - 1;    char tmp = 0;    while (steps--)    {        tmp = str[len];        for (j = len - 1; j >= 0; j--)        {            str[j + 1] = str[j];        }        str[0] = tmp;    }}void test1(){    char p[] = "abcdef";    rightloopmove(p, 2);    printf("%s\n", p);}enum State{    legal,    illegal};enum State state = illegal;int my_atoi(const char *str){    assert(str);    int flag = 1;    long long ret = 0;    if (*str == '\0')    {        return 0;    }    while (isspace(*str))    {        str++;    }    if (*str == '+')    {        str++;    }    if (*str == '-')    {        flag = -1;        str++;    }    while ((*str!='\0')&&(isdigit(*str)))    {        state = legal;        ret = ret * 10 + flag*(*str - '0');        if ((ret > INT_MAX) || (ret < INT_MIN))        {            state = illegal;            return 0;        }        str++;    }    if ((*str!='\0')&&!(isdigit(*str)))    {        state = illegal;    }    return (int)ret;}void test2(){    int ret = my_atoi("       -123");    if (state == legal)    {        printf("%d\n", ret);    }}#define tableSize 256 //创建一个哈希表，因为ASCII码表中只有0~255共256个字符。char First_Char(char* pString,int num){    assert(pString);     int hashTable[tableSize] = {0};    //确定字符串中每个字符出现的次数    char* pHashKey = pString;    while (*(pHashKey) != '\0')    {        hashTable[*(pHashKey++)]++;    }    //找到字符串中只出现k次的那个字符    pHashKey = pString;    while (*pHashKey != '\0')    {        if (hashTable[*pHashKey] == num)            return*pHashKey;        pHashKey++;    }    //如果这个字符串为空，或者字符串中的每个字符都至少出现k+1次    return 0;}void test3(){    char str[1000];  //这个函数是在字符串特别大时建议使用，故定义一个大小为1000的数组    printf("请输入字符串：");    gets(str);    int k = 3;    if (First_Char(str,k) == 0)        printf("输入字符串中没有找到第一个只出现k次的字符！\n");    else        printf("输入字符串中第一个只出现k次的字符为：%c\n", First_Char(str,k));}int reverse(int num){    int ret = 0;    while (num)    {        ret = ret * 10 + num % 10;        num = num / 10;    }    return ret;}void test4(){    int num = -123;    printf("%d\n",reverse(num));}void  leftloopmove(int *str, unsigned short steps,int sz){    assert(str != NULL);    int i = 0;    int j = 0;    int len = sz - 1;    int tmp = 0;    while (steps--)    {        tmp = str[0];        for (j = 0; j < len; j++)        {            str[j] = str[j+1];        }        str[len] = tmp;    }}int New_Binary_Search(int arr[], int num, int sz)//使用二分查找，每次都把数组分成两半，判别哪一半是排过序的。//如果待查元素在排过序的那一半，那么用二分查找。//如果在另一半，那么我们递归地在这一半里继续查找。{    int left = 0;    int right = sz - 1;    while (left <= right)    {        int mid = left + (right - left) / 2;        if (arr[mid] == num)        {            return mid;        }        if (arr[left] <= arr[mid])        { // 左边是排好序的              if (arr[left] <= num && num < arr[mid])            {                right = mid - 1;            }            else            {                left = mid + 1;            }        }        else         {            if (arr[mid] < num && num <= arr[right])            {                left = mid + 1;            }            else            {                right = mid - 1;            }        }    }    return -1;}void test5(){    int arr[] = { 3, 4, 5, 6, 7,8, 9, 1, 2 };    int i = 0;    int sz = (sizeof(arr) / sizeof(arr[0]));    //leftloopmove(arr, 2, sz);    for (i = 0; i < sz; i++)    {        printf("%d",arr[i]);    }    printf("\n");    int ret = New_Binary_Search(arr, 1, sz);    if (ret != -1)    {        printf("下标为%d",ret);    }    else    {        printf("没找到！");    }}int main(){    //test();////指定位置1或置0    //test1();//右旋    //test2();//模拟实现atoi    //test3();//首次出现K次    //test4();//翻转整型    test5();//旋转已排序数组查找数字    system("pause");    return 0;}