查询两个数组的中位数

Median of Two Sorted Arrays

• 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

思路

1. 给定nums1给nums2，则中位数的位置一定能确定，len(nums1) + len(nums2) / 2的取整值，记为median_idx
2. 需要区分是len(nums1) + len(nums2)是奇数还是偶数
3. 两个指针，只需要从nums1和nums2头部开始往后移动，将两者中较小值放入新的数组
4. 当新的数组的长度达到了median_idx，即可求中位数
5. 过程中需要注意两个数组长度不等的情况，防止下标越界

代码

class Solution(object):    def findMedianSortedArrays(self, nums1, nums2):        """        :type nums1: List[int]        :type nums2: List[int]        :rtype: float        """        median_idx = int((len(nums1) + len(nums2)) / 2)        index1, index2 = 0, len(nums1)        combine_list = nums1 + nums2        new_list = []        i = 0        while i <= median_idx:            if index1 < len(nums1) and index2 - len(nums1) < len(nums2):                new_list.append(min(combine_list[index1], combine_list[index2]))                if combine_list[index1] < combine_list[index2]:                    index1 += 1                else:                    index2 += 1            elif index1 >= len(nums1):                # nums1 is out of range, append nums2                new_list.append(combine_list[index2])                index2 += 1            elif index2 >= len(nums2):                # nums2 is out of range, append nums1                new_list.append(combine_list[index1])                index1 += 1            i += 1        if (len(nums1) + len(nums2)) % 2 == 0:            median = float(new_list[median_idx] + new_list[median_idx - 1]) / 2        else:            median = new_list[median_idx]        return median

本题以及其它leetcode题目代码github地址: github地址