c语言二分查找的实现

二分查找也称折半查找（Binary Search），它是一种效率较高的查找方法。但是，折半查找要求线性表必须采用顺序存储结构，而且表中元素按关键字有序排列。

通常在查找数组的时候，会发现一个一个查找很浪费资源，因此通过二分查找法实现 。

算法要求：

1.必须采用顺序存储结构。

2.必须按关键字大小有序排列。

首先，假设表中元素是按升序排列，将表中间位置记录的关键字与查找关键字比较，如果两者相等，则查找成功；否则利用中间位置记录将表分成前、后两个子表，如果中间位置记录的关键字大于查找关键字，则进一步查找前一子表，否则进一步查找后一子表。重复以上过程，直到找到满足条件的记录，使查找成功，或直到子表不存在为止，此时查找不成功。

（上面是思维导图）

#include#include //while、do while 、goto、for、递归void digui(int shang,int xia,int zhong,int num){    zhong = (shang + xia) / 2;    if (shang <= xia)    {        if (num == zhong)        {            printf("递归:find.%d\n", zhong);            return;        }        else if (num < zhong)        {            xia = zhong - 1;            printf("shang.%d,zhong.%d,xia.%d \n", shang, zhong, xia);        }        else        {            shang = zhong + 1;            printf("shang.%d,zhong.%d,xia.%d \n", shang, zhong, xia);        }    }    return;}int main(){    int a[1024];    for (int i = 0; i<1024; i++) {        a[i] = i;        //printf("%d\n",a[i]);    }    int shang = 0;    int xia = 1023;    int zhong = (shang + xia) / 2;    int num = 102;    while (shang <= xia)//一、while    {        zhong = (shang + xia) / 2;        if (num == zhong)        {            printf("while:find.%d\n", zhong);            break;        }        else if (num < zhong)        {            xia = zhong - 1;            printf("shang.%d,zhong.%d,xia.%d \n", shang, zhong, xia);        }        else        {            shang = zhong + 1;            printf("shang.%d,zhong.%d,xia.%d \n", shang, zhong, xia);        }    }    printf("==================\n");    printf("==================\n");    printf("==================\n");    shang = 0;    xia = 1023;    zhong = (shang + xia) / 2;    num = 102;    do {//二、do while        zhong = (shang + xia) / 2;        if (num == zhong)        {            printf("do while:find.%d\n", zhong);            break;        }        else if (num < zhong)        {            xia = zhong - 1;            printf("shang.%d,zhong.%d,xia.%d \n", shang, zhong, xia);        }        else        {            shang = zhong + 1;            printf("shang.%d,zhong.%d,xia.%d \n", shang, zhong, xia);        }    } while (xia>=shang);    printf("==================\n");    printf("==================\n");    printf("==================\n");    shang = 0;    xia = 1023;    zhong = (shang + xia) / 2;    num = 988;AAA://三、goto    zhong = (xia + shang) / 2;    if (num == zhong)    {        printf("goto:find.%d\n", zhong);        goto BBB;    }    else if (num > zhong)    {        shang = zhong + 1;        printf("shang.%d,zhong.%d,xia.%d \n", shang, zhong, xia);    }    else    {        xia = zhong - 1;        printf("shang.%d,zhong.%d,xia.%d \n", shang, zhong, xia);    }    if (shang<=xia)    {        goto AAA;    }BBB:    printf("==================\n");    printf("==================\n");    printf("==================\n");    shang = 0;    xia = 1023;    zhong = (shang + xia) / 2;    num = 288;    for (int i; shang <= xia ;zhong = (shang + xia) / 2)    {        if (num == zhong)        {            printf("for:find.%d\n", zhong);            break;        }        else if (num < zhong)        {            xia = zhong - 1;            printf("shang.%d,zhong.%d,xia.%d \n", shang, zhong, xia);        }        else        {            shang = zhong + 1;            printf("shang.%d,zhong.%d,xia.%d \n", shang, zhong, xia);        }    }    printf("==================\n");    printf("==================\n");    printf("==================\n");    shang = 0;    xia = 1023;    zhong = (shang + xia) / 2;    num = 288;    digui(shang,xia,zhong,num); //递归    getchar();    getchar();     Sleep(5000);    return 0;}