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.

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.
Answers within 10^-5 of the actual value will be accepted as correct.

class Solution {
    public double[] medianSlidingWindow(int[] nums, int k) {
        PriorityQueue<Integer> maxheap = new PriorityQueue<>((a, b) -> b - a);
        PriorityQueue<Integer> minheap = new PriorityQueue();
        int l = 0, count = 0, ind = 0;
        double[] res = new double[nums.length - k + 1]; 
        for(int r = 0; r < nums.length; r++) {
            if(maxheap.size() == 0 || maxheap.peek() >= nums[r]) maxheap.offer(nums[r]);
            else minheap.offer(nums[r]);
            
            balance(maxheap, minheap);
            
            if(r -l + 1 > k) {
                if(maxheap.contains(nums[l])) {
                    maxheap.remove(nums[l]);
                }
                else minheap.remove(nums[l]);
                l++;
                balance(maxheap, minheap);
            }
            if(r - l + 1 == k) {
                res[ind++] = getmedian(maxheap, minheap);
            }
        }
        return res;
    }
    
    public double getmedian(PriorityQueue<Integer> maxheap, PriorityQueue<Integer> minheap) {
        if(maxheap.size() == minheap.size()) return ((long)maxheap.peek() + minheap.peek()) * 0.5;
        else return (double)maxheap.peek();
    }
    
    public void balance(PriorityQueue<Integer> maxheap, PriorityQueue<Integer> minheap) {
        if(maxheap.size() - 1 > minheap.size()) minheap.offer(maxheap.poll());
        else if(maxheap.size() < minheap.size()) maxheap.offer(minheap.poll());
    }
}

sliding window + max/min heap 40/42 passed, 直觉告诉我是remove那里出了问题，

草草草草草，结果你猜是什么原因？如果不是跑了别人的答案，我永远不会知道错误杵在这里

PriorityQueue<Integer> maxheap = new PriorityQueue<>((a, b) -> b - a);

我？？？ 结果只要换成PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());就可以了

我大概知道了，(a, b) -> b - a在b很大a很小的时候可能会溢出？因为换成(x, y) -> Integer.compare(y, x)又没事了,牛逼了

class Solution {
    public double[] medianSlidingWindow(int[] nums, int k) {
        PriorityQueue<Integer> maxheap = new PriorityQueue<>(Collections.reverseOrder());
        PriorityQueue<Integer> minheap = new PriorityQueue();
        int l = 0, count = 0, ind = 0;
        double[] res = new double[nums.length - k + 1]; 
        for(int r = 0; r < nums.length; r++) {
            if(maxheap.size() == 0 || maxheap.peek() > nums[r]) maxheap.offer(nums[r]);
            else minheap.offer(nums[r]);
            
            balance(maxheap, minheap);
            
            if(r -l + 1 > k) {
                if(maxheap.contains(nums[l])) {
                    maxheap.remove(nums[l]);
                }
                else minheap.remove(nums[l]);
                l++;
                balance(maxheap, minheap);
            }
            if(r - l + 1 == k) {
                res[ind++] = getmedian(maxheap, minheap);
            }
        }
        return res;
    }
    
    public double getmedian(PriorityQueue<Integer> maxheap, PriorityQueue<Integer> minheap) {
        if(maxheap.size() == minheap.size()) return ((long)maxheap.peek() + minheap.peek()) * 0.5;
        else return maxheap.peek();
    }
    
    public void balance(PriorityQueue<Integer> maxheap, PriorityQueue<Integer> minheap) {
        if(maxheap.size() - 1 > minheap.size()) minheap.offer(maxheap.poll());
        else if(maxheap.size() < minheap.size()) maxheap.offer(minheap.poll());
    }
}