数据结构入门-单调队列

原理介绍

双向队列

思考一下：对于数组nums，我们想知道max(nums[i-k],...,nums[i])如何高效处理？

单调队列

单调队列，即从队首到队尾单调的队列。

假设nums = [10,8,6,4,5,12],k=2，单调队列滑动窗口入队出队过程：

需要注意的是在滑动窗口时，我们需要知道原数据节点的下标，所以需要以二元组绑定索引值的形式处理。这种情况下，队首始终是最大值，并满足从队首到队尾单调递减。

请添加图片描述

例题练习

力扣 239. 滑动窗口最大值

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        int n = nums.length;
		int[] ans = new int[n-k+1];
		Deque<Integer[]> ql = new ArrayDeque<Integer[]>();
		for (int i = 0; i < n; i++) {
			while (!ql.isEmpty() && i-ql.peekFirst()[0] >= k) ql.pollFirst();
			Integer[] t = {i, nums[i]};
			while (!ql.isEmpty() && ql.peekLast()[1] < nums[i]) ql.pollLast();
            ql.offerLast(t);
			if (i-k+1 >= 0) ans[i-k+1] = ql.peekFirst()[1];
		}
		return ans;
    }
}

P1886 滑动窗口 /【模板】单调队列

70分~，这道题还是建议使用C++，否则很容易内存超出（MLE）或者时间超出（TLE）。

import java.io.*;
import java.util.ArrayDeque;
import java.util.Deque;
import java.util.Scanner;

public class Main {
	static int[] minSlidingWindow(int[] nums, int k) {
		int n = nums.length;
		int[] ans = new int[n-k+1];
		Deque<Integer[]> ql = new ArrayDeque<Integer[]>();
		for (int i = 0; i < n; i++) {
			while (!ql.isEmpty() && i-ql.peekFirst()[0] >= k) ql.pollFirst();
			Integer[] t = {i, nums[i]};
			while (!ql.isEmpty() && ql.peekLast()[1] > nums[i]) ql.pollLast();
			ql.offerLast(t);
			if (i-k+1 >= 0) ans[i-k+1] = ql.peekFirst()[1];
		}
		return ans;
	}
	static int[] maxSlidingWindow(int[] nums, int k) {
		int n = nums.length;
		int[] ans = new int[n-k+1];
		Deque<Integer[]> ql = new ArrayDeque<Integer[]>();
		for (int i = 0; i < n; i++) {
			while (!ql.isEmpty() && i-ql.peekFirst()[0] >= k) ql.pollFirst();
			Integer[] t = {i, nums[i]};
			while (!ql.isEmpty() && ql.peekLast()[1] < nums[i]) ql.pollLast();
			ql.offerLast(t);
			if (i-k+1 >= 0) ans[i-k+1] = ql.peekFirst()[1];
		}
		return ans;
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int n = sc.nextInt();
		int k = sc.nextInt();
		int[] nums = new int[n];
		for (int i = 0; i < n; i++) {
			nums[i] = sc.nextInt();
		}
		int[] ans1 = minSlidingWindow(nums, k);
		int[] ans2 = maxSlidingWindow(nums, k);
		for (int i = 0; i < n-k+1; i++) {
			out.print(ans1[i]);
			if (i != n-k) out.print(" ");
		}
		out.println();
		out.flush();
		for (int i = 0; i < n-k+1; i++) {
			out.print(ans2[i]);
			if (i != n-k) out.print(" ");
		}
		out.flush();
	}
}