The sword refers to Offer-the maximum value of the sliding window

Topic description

Given an array and the size of the sliding window, find the maximum value of all values ​​in the sliding window. For example, if the input array is {2,3,4,2,6,2,5,1} and the size of the sliding window is 3, then there are 6 sliding windows in total, and their maximum values ​​are {4,4,6, 6,6,5}; There are six sliding windows for the array {2,3,4,2,6,2,5,1}: {[2,3,4],2,6,2,5 ,1}, {2,[3,4,2],6,2,5,1}, {2,3,[4,2,6],2,5,1}, {2,3,4 ,[2,6,2],5,1}, {2,3,4,2,[6,2,5],1}, {2,3,4,2,6,[2,5, 1]}.

ideas

Idea one:

Use a deque to store the indices of array elements

1. If the new value is less than the number at the end of the queue, append it to the end, because it may become the maximum value after the previous maximum value has been windowed
2. If the new value is larger than the tail, delete the tail and append it to the back
3. If the index of the added value is greater than the index of the value at the head of the queue and the index of the value at the head of the queue exceeds the window size, delete the value at the head
4. Each time the head of the queue is the largest value in the sliding window

Idea two:

max heap method

Build a maximum heap with the size of the window, and take the maximum value of the window from the heap each time. As the window slides to the right, it is necessary to delete the top elements in the heap that do not belong to the window.

Code

package StackAndQueue;

import java.util.ArrayList;
import java.util.Deque;
import java.util.LinkedList;
import java.util.PriorityQueue;

/**
 * 滑动窗口的最大值
 * 给定一个数组和滑动窗口的大小，找出所有滑动窗口里数值的最大值。例如，如果输入数组{2,3,4,2,6,2,5,1}及滑动窗口的大小3，那么一共存在6个滑动窗口，他们的最大值分别为{4,4,6,6,6,5}； 针对数组{2,3,4,2,6,2,5,1}的滑动窗口有以下6个： {[2,3,4],2,6,2,5,1}， {2,[3,4,2],6,2,5,1}， {2,3,[4,2,6],2,5,1}， {2,3,4,[2,6,2],5,1}， {2,3,4,2,[6,2,5],1}， {2,3,4,2,6,[2,5,1]}。
 */
public class Solution52 {
    public static void main(String[] args) {
        Solution52 solution52 = new Solution52();
        int[] num = {2, 3, 4, 2, 6, 2, 5, 1};
        int size = 3;
        ArrayList<Integer> list = solution52.maxInWindows(num, size);
        System.out.println(list);

    }

    /**
     * 最大堆方法
     * 构建一个窗口size大小的最大堆，每次从堆中取出窗口的最大值，随着窗口往右滑动，需要将堆中不属于窗口的堆顶元素删除。
     *
     * @param num
     * @param size
     * @return
     */
    public ArrayList<Integer> maxInWindows_2(int[] num, int size) {
        ArrayList<Integer> res = new ArrayList<>();
        if (size > num.length || size < 1) return res;
        // 构建最大堆，即堆顶元素是堆的最大值。
        PriorityQueue<Integer> heap = new PriorityQueue<Integer>((o1, o2) -> o2 - o1);
        for (int i = 0; i < size; i++) heap.add(num[i]);
        res.add(heap.peek());
        for (int i = 1; i + size - 1 < num.length; i++) {
            heap.remove(num[i - 1]);
            heap.add(num[i + size - 1]);
            res.add(heap.peek());
        }
        return res;
    }

    /**
     * 双队列方法
     * 滑动窗口的最大值总是保存在队列首部，队列里面的数据总是从大到小排列。
     *
     * @param num
     * @param size
     * @return
     */
    public ArrayList<Integer> maxInWindows(int[] num, int size) {
        ArrayList<Integer> res = new ArrayList<>();
        if (num == null || num.length == 0 || size == 0 || size > num.length) {
            return res;
        }
        Deque<Integer> deque = new LinkedList<>();
        for (int i = 0; i < num.length; i++) {
            if (!deque.isEmpty()) {
                // 如果队列头元素不在滑动窗口中了，就删除头元素
                if (i >= deque.peek() + size) {
                    deque.pop();
                }

                // 如果当前数字大于队列尾，则删除队列尾，直到当前数字小于等于队列尾，或者队列空
                while (!deque.isEmpty() && num[i] >= num[deque.getLast()]) {
                    deque.removeLast();
                }
            }
            deque.offer(i); // 入队列

            // 滑动窗口经过一个滑动窗口的大小，就获取当前的最大值，也就是队列的头元素
            if (i + 1 >= size) {
                res.add(num[deque.peek()]);
            }
        }
        return res;
    }
}