数据结构入门-单调队列

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。

单调性

单调递增

单调递减

单调不增

单调不减

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。

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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;
    }
}

Go语言手写数据结构

// 链表节点
type Node struct {
	Idx   int
	Val   int
	Left  *Node
	Right *Node
}

// 双向链表
type Deque struct {
	First *Node
	Last  *Node
}

// 判断队空
func (this Deque) IsEmpty() bool {
	return this.First == nil
}

// 队首值
func (this Deque) getFirst() int {
	return this.First.Val
}

// 队尾值
func (this Deque) getLast() int {
	return this.Last.Val
}

// 队首下标
func (this Deque) FirstIndex() int {
	return this.First.Idx
}

// 队尾下标
func (this Deque) LastIndex() int {
	return this.Last.Idx
}

// 队首入队
func (this *Deque) OfferFirst(idx int, val int) bool {
	if this.IsEmpty() {
		this.First = &Node{Idx: idx, Val: val}
		this.Last = this.First
		return true
	}
	this.First.Left = &Node{Idx: idx, Val: val, Right: this.First}
    this.First = this.First.Left
	return true
}

// 队尾入队
func (this *Deque) OfferLast(idx int, val int) bool {
	if this.IsEmpty() {
		this.First = &Node{Idx: idx, Val: val}
		this.Last = this.First
		return true
	}
	this.Last.Right = &Node{Idx: idx, Val: val, Left: this.Last}
    this.Last = this.Last.Right
	return true
}

// 队首出队
func (this *Deque) PollFirst() *Node {
	if this.IsEmpty() {
		return nil
	}
	res := this.First
	if this.First == this.Last {
		this.First = nil
		this.Last = nil
		return res
	}
	this.First = this.First.Right
	this.First.Left = nil
	return res
}

// 队尾出队
func (this *Deque) PollLast() *Node {
	if this.IsEmpty() {
		return nil
	}
	res := this.Last
	if this.First == this.Last {
		this.First = nil
		this.Last = nil
		return res
	}
	this.Last = this.Last.Left
	this.Last.Right = nil
	return res
}

func maxSlidingWindow(nums []int, k int) []int {
	n := len(nums)
	ans := make([]int, n-k+1)
	var ql Deque
	for i, v := range nums {
        // 维护窗口大小
		for !ql.IsEmpty() && i-ql.FirstIndex() >= k {
			ql.PollFirst()
		}
        // 维护单调队列
		for !ql.IsEmpty() && ql.getLast() < v {
			ql.PollLast()
		}
		ql.OfferLast(i, v)
		if i-k+1 >= 0 {
			ans[i-k+1] = ql.getFirst()
		}
	}
	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();
	}
}