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)).

思路：将原问题转变成找两个升序数组第k小数的问题。

public class Solution {
        public double findMedianSortedArrays(int[] nums1, int[] nums2) {
            int len = nums1.length + nums2.length;
            if((len&1) == 1) {
                return findKth(nums1, 0, nums1.length-1, nums2, 0, nums2.length-1, len / 2 + 1);
            } else {
                return findKth(nums1, 0, nums1.length-1, nums2, 0, nums2.length-1, len / 2) / 2 + 
                findKth(nums1, 0, nums1.length-1, nums2, 0, nums2.length-1, len / 2 + 1) / 2;
            }
        }
        
        public double findKth(int[] nums1, int left1, int right1, int[] nums2, int left2, int right2, int k) {
            int len1 = right1 - left1 + 1;
            int len2 = right2 - left2 + 1;
            //always assume that m is equal or smaller than n
            if(len1 > len2) return findKth(nums2, left2, right2, nums1, left1, right1, k);
            if(len1 == 0) return nums2[left2 + k -1];
            if(k == 1) return Math.min(nums1[left1], nums2[left2]);
            //divide k into two parts 
            int part1 = Math.min(k/2, len1);
            int part2 = k - part1;
            int index1 = left1 + part1 - 1;
            int index2 = left2 + part2 - 1;
            if(nums1[index1] < nums2[index2]) return findKth(nums1, index1+1, right1, nums2, left2, right2, k-part1);
            else if(nums1[index1] > nums2[index2]) return findKth(nums1, left1, right1, nums2, index2+1, right2, k-part2);
            else return nums1[index1];
        }
    }

 在最好情况下，每次都有k一半的元素被删除，所以算法复杂度为logk，由于求中位数时k为（m+n）/2，所以算法复杂度为log(m+n)。