LeetCode | Median of Two Sorted Arrays

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;        }    }};