LeetCode 480. Sliding Window Median

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本文介绍了一种计算滑动窗口中位数的算法，该算法通过不断更新窗口内的数值并排序来获取每个窗口的中位数。适用于处理数组中滑动窗口问题，返回每个窗口对应的中位数数组。

Problem Statement

(Source) Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.

Examples:

[2,3,4] , the median is 3

[2,3], the median is (2 + 3) / 2 = 2.5

Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Your job is to output the median array for each window in the original array.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

Window position                Median
---------------               -----
[1  3  -1] -3  5  3  6  7       1
 1 [3  -1  -3] 5  3  6  7       -1
 1  3 [-1  -3  5] 3  6  7       -1
 1  3  -1 [-3  5  3] 6  7       3
 1  3  -1  -3 [5  3  6] 7       5
 1  3  -1  -3  5 [3  6  7]      6

Therefore, return the median sliding window as [1,-1,-1,3,5,6].

Note:
You may assume k is always valid, ie: 1 ≤ k ≤ input array’s size for non-empty array.

Solution

class Solution(object):
    def medianSlidingWindow(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[float]
        """
        n = len(nums)
        res = [0] * (n - k + 1)
        arr = None
        for i in xrange(n-k+1):
            if arr is None: 
                arr = nums[i : i+k]
            else:
                arr.remove(nums[i-1])
                arr.append(nums[i+k-1])
            arr.sort()
            res[i] = arr[k>>1] / 1.0
            if k & 1 == 0:
                res[i] = (arr[k>>1] + arr[-1 + k>>1]) / 2.0
        return res