** Median of Two Sorted Arrays

There are two sorted arrays nums1 and nums2 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 len = A.length + B.length;
        if (len % 2 == 1) {
            return findKth(A, 0, B, 0, len / 2 + 1);
        }
        return (
            findKth(A, 0, B, 0, len / 2) + findKth(A, 0, B, 0, len / 2 + 1)
        ) / 2.0;
    }

    // find kth number of two sorted array
    public static int findKth(int[] A, int A_start,
                              int[] B, int B_start,
                              int k){        
        if (A_start >= A.length) {
            return B[B_start + k - 1];
        }
        if (B_start >= B.length) {
            return A[A_start + k - 1];
        }

        if (k == 1) {
            return Math.min(A[A_start], B[B_start]);
        }
        
        int A_key = A_start + k / 2 - 1 < A.length
                    ? A[A_start + k / 2 - 1]
                    : Integer.MAX_VALUE;
        int B_key = B_start + k / 2 - 1 < B.length
                    ? B[B_start + k / 2 - 1]
                    : Integer.MAX_VALUE; 
        
/*首先假设数组A和B的元素个数都大于k/2，我们比较A[k/2-1]和B[k/2-1]两个元素，这两个元素分别表示A的第k/2小的元素和B的第k/2小的元素。这两个元素比较共有三种情况：>、<和=。如果A[k/2-1]<B[k/2-1]，这表示A[0]到A[k/2-1]的元素都在A和B合并之后的前k小的元素中。换句话说，A[k/2-1]不可能大于两数组合并之后的第k小值，所以我们可以将其抛弃。*/


        if (A_key < B_key) {
            return findKth(A, A_start + k / 2, B, B_start, k - k / 2);
        } else {
            return findKth(A, A_start, B, B_start + k / 2, k - k / 2);
        }
    }
}

reference:

http://blog.csdn.net/yutianzuijin/article/details/11499917