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)).

public class Solution {    public double findMedianSortedArrays(int A[], int B[]) {        int k = A.length + B.length;        return k % 2 == 0 ?  (findK(A, 0, A.length - 1, B, 0, B.length - 1, k/2 + 1) +         findK(A, 0, A.length - 1, B, 0, B.length - 1, k/2)) / 2        : findK(A, 0, A.length - 1, B, 0, B.length - 1, k/2 + 1);    }    //返回两个数组中第k大的元素。    public double findK(int a[], int s1, int e1, int b[], int s2, int e2, int k) {        int m = e1 - s1 + 1;        int n = e2 - s2 + 1;        if (m > n) return findK(b, s2, e2, a, s1, e1, k); //a的长度比b的小。

if (s1 > e1) return b[s2 + k - 1]; if (s2 > e2) return a[s1 + k - 1]; if (k == 1) return Math.min(a[s1], b[s2]); int midA = Math.min(k/2, m), midB = k - midA; //假设a的第midA大的元素比b的第midB大的元素小， //那么删掉a的前midA个元素，在剩余的数中找第k-midA大的。 if (a[s1 + midA - 1] < b[s2 + midB - 1]) return findK(a, s1 + midA, e1, b, s2, e2, k - midA); else if (a[s1 + midA - 1] > b[s2 + midB - 1]) return findK(a, s1, e1, b, s2 + midB, e2, k - midB); else return a[s1 + midA - 1]; } }