题目：

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：

基本思路是从两个数组中依次移除最小最大的一对数，重复操作直到两个数组共剩下1个或者2个数，即为中位数。

代码：

class Solution {
public:
    double findMedianSortedArrays(int A[], int m, int B[], int n) {
        int left_A = 0;
        int right_A = m-1;
        int left_B = 0;
        int right_B = n-1;
        while((right_A>=left_A||right_B>=left_B)&&(getNum(left_A,right_A)+getNum(left_B,right_B))>2)
        {
            getMinVal(left_A,right_A,A)<getMinVal(left_B,right_B,B)?left_A++:left_B++;
            getMaxVal(left_A,right_A,A)<getMaxVal(left_B,right_B,B)?right_B--:right_A--;
        };
        if(left_A==right_A&&left_B==right_B)
        {
            return (A[left_A]+B[left_B])/2.0;
        }
        else if(left_A==right_A)
        {
            return A[left_A];
        }
        else if(left_B==right_B)
        {
            return B[left_B];
        }
        else if(left_A+1==right_A)
        {
            return (A[left_A]+A[right_A])/2.0;
        }
        else if(left_B+1==right_B)
        {
            return (B[left_B]+B[right_B])/2.0;
        }
    }
    
    int getMinVal(int left, int right, int array[])
    {
        if(left>right)
            return INT_MAX;
        else
            return array[left];
    }
    
    int getMaxVal(int left, int right, int array[])
    {
        if(left>right)
            return INT_MIN;
        else
            return array[right];
    }
    
    int getNum(int left, int right)
    {
        if(left>right)
        {
            return 0;
        }
        else
        {
            return right-left+1;
        }
    }
};