4. Median of Two Sorted Arrays

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

• 题目描述

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

• 题目大意

        给出两个排序链表，给出两个链表合并后的中位数，时间复杂度应该为  O(log (m+n))。

•  分析

        这道题的难点在于时间复杂度的要求。直接想出时间复杂度为O(log (m+n))的算法有点难度，先尝试一种简单的方法：

            把两个链表连接起来，再排序，输出中位数即可，主要时间消耗在排序上，网上查阅可知链表的排序算法时间复杂度为O((m+n)*log(m+n)),虽然没有符合题目意思，也Accepted了。在Discuss模块中有大神提供了一种O(log (min(m,n)))的方法https://leetcode.com/problems/median-of-two-sorted-arrays/discuss/2481/Share-my-O(log(min(mn))-solution-with-explanation

• 我的解法

class Solution:
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        num=nums1+nums2
        num.sort()
        # print((num))
        numlen=len(num)
        if (numlen==1):
            return num[0]
        elif (numlen==2):
            return (num[0]+num[1])*0.5
        elif (numlen%2==0):
            return (num[int(numlen/2)]+num[int(numlen/2-1)])*0.5
        else:
            return num[int(numlen/2)]