查找算法系列（一）：二分查找( 描述+代码）

1.分析

1. 需要三个变量
1. nLow
2. nMid
3. nHigh

2. 过程

设待查数据为nFind, 数组为arr    如果 nFind > arr[nMid]         nLow = nMid + 1;    否则        nHigh = nMid - 1;

3. 时间复杂度： O(log^n)

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#include<stdio.h>#include<string.h>int BinaryChop(int* arr, int nLen, int nFind){    if(arr == NULL || nLen <= 0)        return -1;    int nLow = 0;    int nHight = nLen -1 ;    int nMid = (nLow + nHight)/2;    while(nLow <= nHight)    {        if(nFind == arr[nMid])        {            return nMid;                    }        else if(nFind < arr[nMid])        {            nHight = nMid - 1;        }        else        {            nLow = nMid + 1;        }        nMid = (nLow + nHight)/2;    }    return -1;}void Print(int arr[], int len){    for(int i=0; i< len;i++)    {        printf("%d  ",arr[i]);    }    printf("\n");}void test(int* arr, int nLen, int nFind, int except){       int res = BinaryChop(arr,nLen,nFind);    if(except == res)    {        printf("Binary chop: %d is find in %d \n", nFind, res);    }    else    {        printf("Binary chop: %d is not in arr,res = %d \n",nFind,res);    }}int main(){    int arr[] = {1,2,3,5,6,9,11,13,15,17,19,23,45,63,79,80,91,101};    int nLen = sizeof(arr)/sizeof(arr[0]);    Print(arr,nLen);    test(arr, nLen, 10,-1);    test(arr, nLen, 11,6);    test(arr, nLen, 19,10);    test(arr, nLen, 3,2);    return 0;}