C++ 二分查找(折半查找)

时间复杂度:Olog(n)

算法实现如下:

#include<iostream>
using namespace std;

int binarySearch(int arr[],int l,int r,int k);//二分查找 

int main(){
	//二分查找要求数组是有序的 
	int arr[] = {23,54,68,134,145,147,168,190,205,259};
	int idx = binarySearch(arr,0,sizeof(arr)/sizeof(arr[0]) - 1,205);
	cout<<idx<<endl;
	
	return 0;
}

int binarySearch(int arr[],int l,int r,int k){
	if(arr == nullptr)
		return -2;
	int i = l,j = r;
	while(i <= j){
		int mid = (i + j) / 2;
		if(arr[mid] == k)
			return mid;
		
		if(arr[mid] >= k)//用K与中位数进行比较 
		{
			j = mid - 1;//注意这是要mid减去1才行,防止while死循环!!!
		}else{
			i = mid + 1;
		}	
	}
	
	return -1;
}

运行结果如下:


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转载自blog.csdn.net/qq_29762941/article/details/80896433