480. Sliding Window Median

yPhantom 2019年10月24日 31次浏览

原题链接

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]`, 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

这一题给定一个数组以及一个长度为k的窗口,要求窗口中的中位数,并保存为一个数组。

利用大小堆的方法以及滑动窗口,minHeap作为小堆,存放所有大于中位数的数,maxHeap作为大堆,存放所有小于中位数的数,这样中位数的值要不就是小堆的堆顶+大堆的堆顶除以2,要不就是小堆的堆顶。

class Solution {

    private PriorityQueue<Integer> minHeap = new PriorityQueue<>();
    private 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[] result = new double[n];
        for (int i = 0; i <= nums.length; i++) {
            if (i >= k) {
                result[i - k] = getMedian();
                remove(nums[i - k]);
            }
            if (i < nums.length) {
                add(nums[i]);
            }
        }
        return result;
    }
    
    private void add(int num) {
        if (num < getMedian()) {
            maxHeap.offer(num);
        } else {
            minHeap.offer(num);
        }
        
        if (maxHeap.size() > minHeap.size()) {
            minHeap.offer(maxHeap.poll());
        }
        
        if (minHeap.size() - maxHeap.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 (minHeap.size() - maxHeap.size() > 1) {
            maxHeap.offer(minHeap.poll());
        }
        
    }
    
    private double getMedian() {
        if (minHeap.isEmpty() && maxHeap.isEmpty()) {
            return 0;
        }
        if (minHeap.size() == maxHeap.size()) {
            return ((double)minHeap.peek() + (double)maxHeap.peek()) / 2.0;
        } else {
            return (double)minHeap.peek();
        }
    }
}