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);        }    }}