Leetcode||4.Median of Two Sorted Arrays

4. Median of Two Sorted Arrays

• Total Accepted: 122172
• Total Submissions: 604583
• Difficulty: Hard
• Contributors: Admin

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

题意很明确，找中间数，如果为奇数就是中间的那个数，为偶数就是中间两个数的平均数，如果没有要求时间复杂度，最直接的做法是两个数组合并再次排序，但是本题是要求时间复杂度为O(log(m+n))的，网上的做法普遍分两种归并和二分，下面的代码用的归并。

代码如下：

public class Solution {    public double findMedianSortedArrays(int[] nums1, int[] nums2) {        int len1 = nums1.length;        int len2 = nums2.length;        int total = len1 + len2;        if(total % 2==0){            return (findKth(nums1,nums2,total/2)+findKth(nums1,nums2,total/2+1))/2.0;        } else {            return findKth(nums1,nums2,total/2+1);        }    }    private int findKth(int[] nums1, int[] nums2, int k){        int p = 0, q = 0;        for(int i = 0; i < k - 1; i++){            if(p>=nums1.length && q<nums2.length){                q++;            } else if(q>=nums2.length && p<nums1.length){                p++;            } else if(nums1[p]>nums2[q]){                q++;            } else {                p++;            }        }        if(p>=nums1.length) {            return nums2[q];        } else if(q>=nums2.length) {            return nums1[p];        } else {            return Math.min(nums1[p],nums2[q]);        }    }}