一道简单的寻找中位数的题目

**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)).

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

**
但要求是O（log（m+n））
我开始没注意就用了归并排序，竟然过了
amazing。

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        if( nums1.length == 0 && nums2.length == 0){
            double result = 0.0;
            return result;
        }
        int []nums3 = new int[nums1.length + nums2.length];
        int i = 0 ; 
        int j = 0;
        int k = 0;
        while (i < nums1.length && j < nums2.length){
            if (nums1[i] < nums2[j]){
                nums3[k++] = nums1[i++];
            }else {
                nums3[k++] = nums2[j++];
            } 
        }

        while (i < nums1.length){
         nums3[k++] = nums1[i++];
        }

        while (j < nums2.length){
            nums3[k++] = nums2[j++];
        }

        if (nums3.length % 2 == 0){
            double result = ((double)nums3[nums3.length/2] + (double)nums3[nums3.length/2-1])/2;
            return result;
        }
        else{
            double result = nums3[nums3.length/2];
            return result;
        }

    }
}

但是作为努力学习的码畜，当然不能止步于此
所以又去研究别人的算法
Ozx
竟然用递归解决。
高高高，的确是O（log（m+n））
首先判断奇数偶数，递归寻找中位数，最精妙的就是findkth的最后那个k参数，当然两个到零的函数判断返回也是有点难想到，比我的算法好多，佩服佩服

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; 

        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);
        }
    }
}