【数据结构与算法】有序数组二分法查找

一、原理

1、初设low为第一个元素，high为最后一元素，mid由每一次循环 (low+high)/2 计算得出

2、若数组元素个数为偶数（此处不妨设为10个） low=0 ,high=9   则mid=4 （其实mid是4还是5，无所谓）

3、最坏的情况找到（即用了最多的次数找到）

元素个数为奇数时，如图所示 low =3 ，mid =4，high=5。

元素个数为偶数时，如找‘20’这个元素，最后     high=mid-1=3   ，则low=mid=high=3 三者重合

4、若没找到：在low=high下一步，必有low>high，此时循环停止，故得出循环判断条件：low<=high

二、C语言实现

#include <stdio.h>#define M 10//数组长度#define FALSE 0//found的没找到标志#define TRUE 1//found成功找到的标志int binary_search(int a[],int n);int main(void){static int a[M]={-12,0,6,16,23,56,80,100,110,115};int n,found;printf("Input a num to be searched:");scanf("%d",&n);//输入要查找的数found = binary_search(a,n);//返回给foundif(found==-1){printf("There is no such a num\n");}else{printf("The num you find is a[%d]",found);}return 0;}int binary_search(int a[],int n){int low,mid,high,found;low=0;high=M-1;//起始low为0即第一个元素，high为最后一个元素found=FALSE;//预设found为失败状态while (low<=high){mid = (low+high)/2;//（注意是整除）if(n==a[mid])//刚好找到{found=TRUE;break;}else if (n>a[mid])//偏大，本来是low代替mid，但是已知n!=a[mid]，{//故low=mid+1比low=mid再进一步缩小范围low=mid+1;}else//偏小，同理{high=mid-1;}}//若没找到，则low>high，此时found还是==FALSEswitch (found)//判断标志{case TRUE:return mid;case FALSE:return -1;}}

三、递归二分法

#include <stdio.h>#define M 10//数组长度#define FALSE 0//found的没找到标志#define TRUE 1//found成功找到的标志int BinarySearch(int a[],int n, int left, int right);int main(void){static int a[M]={-12,0,6,16,23,56,80,100,110,115};int n,found;printf("Input a num to be searched:");scanf("%d",&n);//输入要查找的数found = BinarySearch(a,n,0,M-1);//返回给foundif(found==-1){printf("There is no such a num\n");}else{printf("The num you find is a[%d]",found);}printf("\n");return 0;}int BinarySearch(int a[],int n, int left, int right) // n为要找的数，left是数组的首元素下标对应low，right是数组的最后一个元素对应high{int mid;if(left<=right){mid = (left+right)/2;if(a[mid]==n) {return mid; //若a[mid]刚好是找要的数，则返回下标}else if(a[mid]<n){left = mid+1;}else if(a[mid]>n){right = mid-1;}return BinarySearch(a,n,left,right);}return -1;//left<=right不满足，则失败}

注意：在else if(a[mid]>n); 处不小心加了分号，在语法上是没有错误的，程序也能执行，但结果是错误的，以后要小心