二分查找(Binary Search, Bin Search)的C语言实现

二分查找需要对数组进行从小到大的排列。
这里展示了递归实现和循环实现2种方法。

#include <stdio.h>

void Bsearch(int list[], int low, int high, int key);
int binSearch(int list[], int n, int key);

void bubble_sort(int list[], int);

int main()
{
    int size, i;
    int list[]={41,67,34,0,69,24,78,58,62,64,5,45,81,27,61,91,95,42,27,36,91,4,2};


    size=sizeof(list)/sizeof(list[0]);
    printf("size=%d\n",size);
    bubble_sort(list, size);
    for(i = 0; i < size; i++)
    {
        printf("%d,", list[i]);
    }

    printf("using Recursion\n");
    Bsearch(list, 0, size, 27);

    printf("Using while loop\n");

    binSearch(list,size,27);
    printf("end\n");

}

void bubble_sort(int list[], int size)
{
    int temp, i, j;
    for (i = 0; i < size; i++)
    {
        for (j = i+1; j < size; j++)
        {
            if (list[i] > list[j])   //list[i] < list[j]
            {
                temp = list[i];
                list[i] = list[j];
                list[j] = temp;
            }
        }
    }
}

void Bsearch(int list[], int low, int high, int key)
{
    int mid;

    if (low > high)
    {
        printf("Key not found\n");
        return;
    }
    mid = (low + high) / 2;
    if (list[mid] == key)
    {
         printf("Key found,mid=%d,list[%d]=%d\n",mid,mid,list[mid]);
         return;
     }
     else if (list[mid] > key)
     {
         Bsearch(list, low, mid - 1, key);
     }
     else if (list[mid] < key)
     {
         Bsearch(list, mid + 1, high, key);
     }
 }



int binSearch(int list[], int n, int key)
{
    //n + 1 个数
    int low = 0;
    int high = n;
    int mid = 0;
    while(low <= high)
    {
        mid = low + ((high-low) >> 1);
        if(key == list[mid])
        {
            printf("mid=%d,list[%d]=%d\n",mid,mid,list[mid]);
            return mid;
        }
        else if(key < list[mid])
        {
            high = mid - 1;
        }
        else
        {
            low = mid + 1;
        }
    }
    return -1;
}

运行结果
这里写图片描述

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转载自blog.csdn.net/ambercctv/article/details/80026530