480. Sliding Window Median

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.

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       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]
      
      
      
      

       
       
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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.

思路：
这题和那道Find Median from Data Stream比起来多加了个sliding window。那道题巧妙的用了两个heap来找到mean，还有道题是Slide Window Maximum，同样是slide window的题。还是用两个heap来做，remove这个操作复杂度用了logk。minHeap和maxHeap，maxHeap在保存较小的一半元素，minHeap保存较大的一半元素，0 <= minHeap.size() - maxHeap.size() <= 1，注意maxheap写的时候不能用a - b，因为可能overflow。

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       class Solution
       
       
    public double[] medianSlidingWindow(int[] nums, int k) {
       
       
        int n = nums.length;
       
       
        double[] result = new double[n - k + 1];
       
       
        maxHeap = new PriorityQueue<>(k/2 + 1, (a, b) -> b.compareTo(a));
       
       
        minHeap = new PriorityQueue<>(k/2 + 1);
       
       

       
       
        for(int i = 0; i < n; i++) {
       
       
            // delete the element beyond the window
       
       
            if(maxHeap.size() + minHeap.size() == k) 
       
       
                slide(nums[i - k]);
       
       
            // add new element to the window
       
       
            add(nums[i]);
       
       
            if(i >= k - 1) {
       
       
                result[i - k + 1] = getMedian();
       
       
            }
       
       
        }
       
       

       
       
        return result;
       
       
    }
       
       

       
       
    PriorityQueue<Integer> minHeap;
       
       
    PriorityQueue<Integer> maxHeap;
       
       
    private void slide(int target) {
       
       
        if(minHeap.contains(target)) 
       
       
            minHeap.remove(target);
       
       
        else maxHeap.remove(target);
       
       
    }
       
       

       
       
    private void add(int num) {
       
       
        maxHeap.add(num);
       
       
        minHeap.add(maxHeap.poll());
       
       
        if(maxHeap.size() + 1 
       
       
            maxHeap.add(minHeap.poll());
       
       
    }
       
       

       
       
    private double getMedian() {
       
       
        // window size is even
       
       
        if(minHeap.size() == maxHeap.size()) 
       
       
            return minHeap.peek()/2.0 + maxHeap.peek()/2.0;
       
       
        else return minHeap.peek();
       
       
    }
       
       
}
      
      
      
      

       
       
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