480. Sliding Window Median

最新推荐文章于 2021-02-27 09:56:42 发布

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最新推荐文章于 2021-02-27 09:56:42 发布

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分类专栏： LeetCode 文章标签： LeetCode Sliding Window Media

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本文介绍了一种高效计算滑动窗口中位数的方法，通过维护最小堆和最大堆来确保每次移动窗口时都能快速获取当前窗口内的中位数。适用于数据流处理等场景。

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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.

这道题可以用堆进行优化，算法复杂度O(nlogk)。维持一个minheap，maxheap，用TreeSet实现。为了防止重复，使用node存储，包含一个val和一个id。当val相同时，返回id的差。这里要考虑Integer.MAX_VALUE和MIN_VALUE的corner case。防溢出的操作一个是在返回(long)maxHeap.first().val + (long)minHeap.last().val) / 2.0时，一个是在TreeSet里的node比较时，不能单纯返回this.val - other.val，会溢出，先判断两者大小，再根据情况返回1或者-1。代码如下：

public class Solution {
    class Node implements Comparable<Node>{
        int val, id;
        public Node(int _val, int _id) {
            this.val = _val;
            this.id = _id;
        }
        public int compareTo(Node other) {
            Node a = (Node) other;
            if (this.val == a.val) {
                return this.id - a.id;
            } else if (this.val > a.val) {
                return 1;
            } else {
                return -1;
            }
        }
    }
    
    public void add(TreeSet<Node> minHeap, TreeSet<Node> maxHeap, Node node) {
        maxHeap.add(node);
        Node min = maxHeap.first();
        minHeap.add(min);
        maxHeap.remove(min);
        while (minHeap.size() > maxHeap.size() + 1) {
            Node max = minHeap.last();
            maxHeap.add(max);
            minHeap.remove(max);
        }
    }
    
    public void remove(TreeSet<Node> minHeap, TreeSet<Node> maxHeap, Node node) {
        if (minHeap.contains(node)) {
            minHeap.remove(node);
        } else {
            maxHeap.remove(node);
        }
    }
    
    public double[] medianSlidingWindow(int[] nums, int k) {
        TreeSet<Node> minHeap = new TreeSet<Node>();
        TreeSet<Node> maxHeap = new TreeSet<Node>();
        double[] res = new double[nums.length - k + 1];
        int index = 0;
        for (int i = 0; i < k - 1; i ++) {
            add(minHeap, maxHeap, new Node(nums[i], i));
            //System.out.print(minHeap.size() + "    ");
            //System.out.println(maxHeap.size());
        }
        for (int i = k - 1; i < nums.length; i ++) {
            add(minHeap, maxHeap, new Node(nums[i], i));
            res[index ++] = minHeap.size() == maxHeap.size()? ((long)maxHeap.first().val + (long)minHeap.last().val) / 2.0: minHeap.last().val;
            remove(minHeap, maxHeap, new Node(nums[i - k + 1], i - k + 1));
            //System.out.print(minHeap.size() + "    ");
            //System.out.println(maxHeap.size());
        }
        return res;
    }
}