文章目录

• 480. 滑动窗口中位数
• • 方法一：双优先队列 + 延迟删除
  • • Solution 类
    • DualHeap 类

480. 滑动窗口中位数

参考leetcode.295. 数据流的中位数

方法一：双优先队列 + 延迟删除

Solution 类

class Solution {
    public double[] medianSlidingWindow(int[] nums, int k) {
        DualHeap dh = new DualHeap(k);
        for (int i = 0; i < k; ++i) {
            dh.insert(nums[i]);
        }
        double[] ans = new double[nums.length - k + 1];
        ans[0] = dh.getMedian();
        for (int i = k; i < nums.length; ++i) {
            dh.insert(nums[i]);
            dh.erase(nums[i - k]);
            ans[i - k + 1] = dh.getMedian();
        }
        return ans;
    }

	 class DualHeap {
	    //...代码放在下面  
	 }
 
}

DualHeap 类

    class DualHeap {
        // 大根堆，维护较小的一半元素
        private PriorityQueue<Integer> small;
        // 小根堆，维护较大的一半元素
        private PriorityQueue<Integer> large;
        // 哈希表，记录「延迟删除」的元素，key 为元素，value 为需要删除的次数
        private Map<Integer, Integer> delayed;

        private int k;
        // small 和 large 当前包含的元素个数，需要扣除被「延迟删除」的元素
        private int smallSize, largeSize;

        public DualHeap(int k) {
            this.small = new PriorityQueue<Integer>(new Comparator<Integer>() {
                public int compare(Integer num1, Integer num2) {
                    return num2.compareTo(num1);
                }
            });
            this.large = new PriorityQueue<Integer>(new Comparator<Integer>() {
                public int compare(Integer num1, Integer num2) {
                    return num1.compareTo(num2);
                }
            });
            this.delayed = new HashMap<Integer, Integer>();
            this.k = k;
            this.smallSize = 0;
            this.largeSize = 0;
        }

        public double getMedian() {
            return (k & 1) == 1 ? small.peek() : ((double) small.peek() + large.peek()) / 2;
        }

        public void insert(int num) {
            if (small.isEmpty() || num <= small.peek()) {
                small.offer(num);
                ++smallSize;
            } else {
                large.offer(num);
                ++largeSize;
            }
            makeBalance();
        }

        public void erase(int num) {
            delayed.put(num, delayed.getOrDefault(num, 0) + 1);
            if (num <= small.peek()) {
                --smallSize;
                if (num == small.peek()) {
                    prune(small);
                }
            } else {
                --largeSize;
                if (num == large.peek()) {
                    prune(large);
                }
            }
            makeBalance();
        }

        // 不断地弹出 heap 的堆顶元素，并且更新哈希表
        private void prune(PriorityQueue<Integer> heap) {
            while (!heap.isEmpty()) {
                int num = heap.peek();
                if (delayed.containsKey(num)) {
                    delayed.put(num, delayed.get(num) - 1);
                    if (delayed.get(num) == 0) {
                        delayed.remove(num);
                    }
                    heap.poll();
                } else {
                    break;
                }
            }
        }

        // 调整 small 和 large 中的元素个数，使得二者的元素个数满足要求
        private void makeBalance() {
            if (smallSize > largeSize + 1) {
                // small 比 large 元素多 2 个
                large.offer(small.poll());
                --smallSize;
                ++largeSize;
                // small 堆顶元素被移除，需要进行 prune
                prune(small);
            } else if (smallSize < largeSize) {
                // large 比 small 元素多 1 个
                small.offer(large.poll());
                ++smallSize;
                --largeSize;
                // large 堆顶元素被移除，需要进行 prune
                prune(large);
            }
        }
    }

标签：int,median,delayed,large,window,num,smallSize,small,leetcode.480
来源： https://blog.csdn.net/e891377/article/details/113622082