[LeetCode] 480. Sliding Window Median

最新推荐文章于 2025-01-02 19:21:53 发布

转载最新推荐文章于 2025-01-02 19:21:53 发布 · 126 阅读

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原文链接：https://segmentfault.com/a/1190000017397584

本文介绍了一种高效计算滑动窗口中位数的算法，通过使用两个堆（最大堆和最小堆）来动态维护窗口内数值的平衡，从而在每次窗口滑动时快速求得中位数。该算法适用于数据流处理和实时数据分析场景。

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Problem

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: k is always smaller than input array's size for non-empty array.

Solution

class Solution {
    //minHeap存大于median的数
    //maxHeap存小于median的数
    PriorityQueue<Integer> minHeap = new PriorityQueue<>();
    PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b)->(b.compareTo(a)));
    public double[] medianSlidingWindow(int[] nums, int k) {
        int n = nums.length-k+1;
        if (n <= 0) return new double[0];
        
        double[] res = new double[n];
        
        for (int i = 0; i <= nums.length; i++) {
            if (i >= k) {
                res[i-k] = getMedian();
                remove(nums[i-k]);
            }
            if (i < nums.length) {
                add(nums[i]);
            }
        }
        
        return res;
    }
    private void add(int num) {
        if (num < getMedian()) maxHeap.offer(num);
        else minHeap.offer(num);
        //保证minHeap总比maxHeap的size相等或多一个元素
        if (maxHeap.size() > minHeap.size()) minHeap.offer(maxHeap.poll());
        if (maxHeap.size() < minHeap.size()-1) maxHeap.offer(minHeap.poll());
    }
    private void remove(int num) {
        if (num < getMedian()) maxHeap.remove(num);
        else minHeap.remove(num);
        if (maxHeap.size() > minHeap.size()) minHeap.offer(maxHeap.poll());
        if (maxHeap.size() < minHeap.size()-1) maxHeap.offer(minHeap.poll());
    }
    private double getMedian() {
        if (maxHeap.isEmpty() && minHeap.isEmpty()) return 0;
        if (maxHeap.size() == minHeap.size()) return ((double) maxHeap.peek() + (double) minHeap.peek()) / 2.0;
        else return (double) minHeap.peek();
    }
}