leetcode: 4. Median of Two Sorted Arrays (java)

题目链接：https://leetcode.com/problems/median-of-two-sorted-arrays/

题目：

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)).

解析：一开始采用了归并排序的思想来解这道题，思想简单，但是复杂度为O(m+n);

后来借鉴了别人的思路：http://blog.csdn.net/linhuanmars/article/details/19905515

java 代码实现：

public class Solution {

    public double findMedianSortedArrays(int[] nums1, int[] nums2) {

        int length = nums1.length + nums2.length;

        if (length%2 == 1){

            return recrusive(nums1, nums2, 0, nums1.length-1, 0, nums2.length-1, length/2 + 1);

        } else {

            return (recrusive(nums1, nums2, 0, nums1.length-1, 0, nums2.length-1, length/2)+recrusive(nums1, nums2, 0, nums1.length-1, 0, nums2.length-1, length/2 + 1))/2.0;

        }

    }

    public double recrusive(int[] A, int[] B, int i, int i2, int j, int j2, int k){

        int m = i2-i+1;

        int n = j2-j+1;

        if (m > n) return recrusive(B, A, j, j2, i, i2, k);

        if (m == 0) {

            return B[j+k-1];

        }

        if (k == 1){

            return Math.min(A[i], B[j]);

        }

        int posA = Math.min(k/2, m);

        int posB = k-posA;

        if(A[i+posA-1] == B[j+posB-1]){

            return A[i+posA-1];

        }else if (A[i+posA-1] < B[j+posB-1]){

            return recrusive(A, B, i+posA, i2, j, j+posB-1, k-posA);

        }else {

            return recrusive(A, B, i, i+posA-1, j+posB, j2, k-posB);

        }

    }

}