循环有序数组查找一个数

女爷i 2022-08-12 01:12 166阅读 0赞

网上看了有些代码有些错误，主要是下面实例中的0查找不到，这里重新更正了下

思想：循环数组有一边是有序的，首先先判断哪一边有序（通过将当前mid点与最左边节点比较），然后查看是否在有序边上

代码如下

#include<iostream>
    #include<assert.h>
    using namespace std;
    int Binary_Search(int *a,int low,int high,int value)
    {
    	int mid=(low+high)/2;
    	if(a[mid] == value)
    	return mid;
        if(a[mid]>=a[low])       //左有序
        {
            if(a[low]<=value && value<a[mid])        //说明value在左边，直接二分查找
            {
                return Binary_Search(a,low,mid,value);
            }
            else                                       //value在右边
            {
                return Binary_Search(a,mid+1,high,value);
            }
        }
        else                    //右有序
        {
            if(a[mid]<=value && value<=a[high])
            {
                return Binary_Search(a,mid,high,value);
            }
            else
            {
                return Binary_Search(a,low,mid-1,value);
            }
    	}
    }
    int main()
    {
    	int A[10] = {7,8,9,0,1,2,3,4,5,6};
    	int p = Binary_Search(A,0,9,0);
    	cout<<p<<endl;
    	return 0;
    }

非递归写法如下：

#include<iostream>
    #include<stdio.h>
    #include<assert.h>
    using namespace std;
    int Binary_Search(int *A,int size,int value)
    {
    	assert(A != NULL);
    	int low = 0;
    	int high = size - 1;
    	while(low <= high)
    	{
    		int mid = (low + high) / 2;
    		if(A[mid] == value)
    			return mid;
    		if(A[low] <= A[mid])//左有序,注意‘=’的使用
    		{
    			if(A[low] <= value && value <= A[mid])
    				high = mid - 1;
    			else 
    				low = mid + 1;
    		}
    		else//右有序
    		{
    			if(A[mid] <= value && A[high] >= value)
    				low = mid + 1;
    			else
    				high = mid - 1;
    		}
    	}
    }
    int main()
    {
    	int A[10] = {7,8,9,0,1,2,3,4,5,6};
    	cout<<Binary_Search(A,10,5);
    	return 0;
    }