Leetcode: 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)).

If we can find kth element of two sorted arrays, then it would be easy to find the median of the two arrays. Use recursion to find kth element of two arrays: use binary search, compare the k/2 th elements in each array, and delete the left part of the smaller one, continue to find k - k/2 th element of the rested arrays, until k = 1, which means that a total number of k - 1 smallest elements are deleted.

public class Solution {    public double findMedianSortedArrays(int A[], int B[]) {        int size = A.length + B.length;        if (size % 2 == 0) {            return (findKth(A, 0, B, 0, size / 2) + findKth(A, 0, B, 0, size / 2 + 1)) / 2.0;        } else {            return findKth(A, 0, B, 0, size / 2 + 1);        }            }        private 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;                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);        }    }}