leetcode 之 Median of Two Sorted Arrays

标签：median of two sorted   中位数   leetcode   递归   

Median of Two Sorted Arrays

There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

思路：首先判断中位数是第几个数，然后把中位数的个数除以2，一半在A中，一半在B中，如果A[mid] < B[mid]，则中位数肯定不在A的mid之前，然后只考虑A的后半段，同时把中位数的个数减少mid，继续递归

class Solution {
public:
int find(int a[],int m,int b[],int n,int index)
{
	if(m==0)return b[index-1];
	if(n==0)return a[index-1];
	if(index==1)return(a[0]<b[0]?a[0]:b[0]);
	int mid = index / 2;//每次把剩下的中位数的个数减半，分到两个数组中考虑
	if(min(m,n)<mid)mid = min(m,n);
	if(a[mid-1] < b[mid-1])return find(a+mid,m-mid,b,n,index-mid);
	else return find(a,m,b+mid,n-mid,index-mid);
}
    double findMedianSortedArrays(int A[], int m, int B[], int n) {
        int mid = (m+n+1)/2;//第几个数是中位数
	double temp = find(A,m,B,n,mid);
	if((m+n)&0x1==1)return temp;
	else return (temp+find(A,m,B,n,mid+1))/2.0;//偶数的情况下，两个中间点数取平均数
    }
};

leetcode 之 Median of Two Sorted Arrays

标签：median of two sorted   中位数   leetcode   递归   

原文地址：http://blog.csdn.net/fangjian1204/article/details/39135507