LeetCode 480. Sliding Window Median 延迟双堆

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本文介绍了一种计算滑动窗口中位数的有效算法。针对整数数组和指定窗口大小，该算法能够准确地找到每个窗口内的中位数，并处理数组元素的变化。通过使用最大堆和最小堆的数据结构来维护窗口内数字的平衡分布。

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: kis always smaller than input array's size for non-empty array.
Answers within 10^-5 of the actual value will be accepted as correct.

-------------------------------

简化代码思路，包函数

bug1: 两个堆用同一个dic

bug2:调整完没有clear_top

class Solution:
    def medianSlidingWindow(self, nums: List[int], k: int) -> List[float]:
        def clear_top(pq, drop_dic, is_min_heap):
            while (pq and drop_dic[pq[0] if is_min_heap else -pq[0]] > 0):
                item = heapq.heappop(pq) if is_min_heap else -heapq.heappop(pq)
                drop_dic[item] -= 1

        def pop_item(pq, drop_dic, is_min_heap):
            if pq:
                res = heapq.heappop(pq) if is_min_heap else -heapq.heappop(pq)
                clear_top(pq, drop_dic, is_min_heap)
                return res

        def push_item(item, pq, drop_dic, is_min_heap):
            element = item if is_min_heap else -item
            heapq.heappush(pq, element)
            clear_top(pq, drop_dic, is_min_heap)

        max_heap, min_heap, max_dic, min_dic, res = [], [], defaultdict(int), defaultdict(int), []
        max_cnt, min_cnt, l = 0, 0, len(nums)
        for i in range(l):
            if (i >= k):
                if (nums[i-k] >= min_heap[0]):
                    min_cnt,min_dic[nums[i-k]] = min_cnt-1,min_dic[nums[i-k]]+1
                    clear_top(min_heap, min_dic, True)
                else:
                    max_cnt,max_dic[nums[i-k]] = max_cnt-1,max_dic[nums[i-k]]+1
                    clear_top(max_heap, max_dic, False)

            push_item(nums[i], min_heap, min_dic, True)
            min_heap_top = pop_item(min_heap, min_dic, True)
            push_item(min_heap_top, max_heap, max_dic, False)
            max_cnt += 1

            if (max_cnt > min_cnt):
                max_heap_top = pop_item(max_heap, max_dic, False)
                push_item(max_heap_top, min_heap, min_dic, True)
                max_cnt, min_cnt = max_cnt-1, min_cnt+1
            if (i >= k - 1):
                res.append((min_heap[0] - max_heap[0]) / 2.0 if k % 2 == 0 else min_heap[0])
        return res