KWAN的解忧杂货铺

# 一.双端队列简介

# 1.概述

双端队列、队列、栈对比

定义 特点
队列 一端删除（头）另一端添加（尾） First In First Out
栈 一端删除和添加（顶） Last In First Out
双端队列 两端都可以删除、添加
优先级队列 优先级高者先出队
延时队列 根据延时时间确定优先级
并发非阻塞队列 队列空或满时不阻塞
并发阻塞队列 队列空时删除阻塞、队列满时添加阻塞

# 2.方法比较

注 1：

  • Java 中 LinkedList 即为典型双端队列实现，不过它同时实现了 Queue 接口，也提供了栈的 push pop 等方法

注 2：

  • 不同语言，操作双端队列的方法命名有所不同，参见下表

    操作 Java JavaScript C++ leetCode 641
    尾部插入 offerLast push push_back insertLast
    头部插入 offerFirst unshift push_front insertFront
    尾部移除 pollLast pop pop_back deleteLast
    头部移除 pollFirst shift pop_front deleteFront
    尾部获取 peekLast at(-1) back getRear
    头部获取 peekFirst at(0) front getFront

  • 吐槽一下 leetCode 命名比较 low

  • 常见的单词还有 enqueue 入队、dequeue 出队

# 3.接口定义

public interface Deque<E> {

    boolean offerFirst(E e);

    boolean offerLast(E e);

    E pollFirst();

    E pollLast();

    E peekFirst();

    E peekLast();

    boolean isEmpty();

    boolean isFull();
}
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# 4.链表实现

/**
 * 基于环形链表的双端队列
 * @param <E> 元素类型
 */
public class LinkedListDeque<E> implements Deque<E>, Iterable<E> {

    @Override
    public boolean offerFirst(E e) {
        if (isFull()) {
            return false;
        }
        size++;
        Node<E> a = sentinel;
        Node<E> b = sentinel.next;
        Node<E> offered = new Node<>(a, e, b);
        a.next = offered;
        b.prev = offered;
        return true;
    }

    @Override
    public boolean offerLast(E e) {
        if (isFull()) {
            return false;
        }
        size++;
        Node<E> a = sentinel.prev;
        Node<E> b = sentinel;
        Node<E> offered = new Node<>(a, e, b);
        a.next = offered;
        b.prev = offered;
        return true;
    }

    @Override
    public E pollFirst() {
        if (isEmpty()) {
            return null;
        }
        Node<E> a = sentinel;
        Node<E> polled = sentinel.next;
        Node<E> b = polled.next;
        a.next = b;
        b.prev = a;
        size--;
        return polled.value;
    }

    @Override
    public E pollLast() {
        if (isEmpty()) {
            return null;
        }
        Node<E> polled = sentinel.prev;
        Node<E> a = polled.prev;
        Node<E> b = sentinel;
        a.next = b;
        b.prev = a;
        size--;
        return polled.value;
    }

    @Override
    public E peekFirst() {
        if (isEmpty()) {
            return null;
        }
        return sentinel.next.value;
    }

    @Override
    public E peekLast() {
        if (isEmpty()) {
            return null;
        }
        return sentinel.prev.value;
    }

    @Override
    public boolean isEmpty() {
        return size == 0;
    }

    @Override
    public boolean isFull() {
        return size == capacity;
    }

    @Override
    public Iterator<E> iterator() {
        return new Iterator<E>() {
            Node<E> p = sentinel.next;
            @Override
            public boolean hasNext() {
                return p != sentinel;
            }

            @Override
            public E next() {
                E value = p.value;
                p = p.next;
                return value;
            }
        };
    }

    static class Node<E> {
        Node<E> prev;
        E value;
        Node<E> next;

        public Node(Node<E> prev, E value, Node<E> next) {
            this.prev = prev;
            this.value = value;
            this.next = next;
        }
    }

    Node<E> sentinel = new Node<>(null, null, null);
    int capacity;
    int size;

    public LinkedListDeque(int capacity) {
        sentinel.next = sentinel;
        sentinel.prev = sentinel;
        this.capacity = capacity;
    }
}
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# 5.数组实现

/**
 * 基于循环数组实现, 特点
 * <ul>
 *     <li>tail 停下来的位置不存储, 会浪费一个位置</li>
 * </ul>
 * @param <E>
 */
public class ArrayDeque1<E> implements Deque<E>, Iterable<E> {

    /*
                    h
            t
        0   1   2   3
        b           a
     */
    @Override
    public boolean offerFirst(E e) {
        if (isFull()) {
            return false;
        }
        head = dec(head, array.length);
        array[head] = e;
        return true;
    }

    @Override
    public boolean offerLast(E e) {
        if (isFull()) {
            return false;
        }
        array[tail] = e;
        tail = inc(tail, array.length);
        return true;
    }

    @Override
    public E pollFirst() {
        if (isEmpty()) {
            return null;
        }
        E e = array[head];
        array[head] = null;
        head = inc(head, array.length);
        return e;
    }

    @Override
    public E pollLast() {
        if (isEmpty()) {
            return null;
        }
        tail = dec(tail, array.length);
        E e = array[tail];
        array[tail] = null;
        return e;
    }

    @Override
    public E peekFirst() {
        if (isEmpty()) {
            return null;
        }
        return array[head];
    }

    @Override
    public E peekLast() {
        if (isEmpty()) {
            return null;
        }
        return array[dec(tail, array.length)];
    }

    @Override
    public boolean isEmpty() {
        return head == tail;
    }

    @Override
    public boolean isFull() {
        if (tail > head) {
            return tail - head == array.length - 1;
        } else if (tail < head) {
            return head - tail == 1;
        } else {
            return false;
        }
    }

    @Override
    public Iterator<E> iterator() {
        return new Iterator<E>() {
            int p = head;
            @Override
            public boolean hasNext() {
                return p != tail;
            }

            @Override
            public E next() {
                E e = array[p];
                p = inc(p, array.length);
                return e;
            }
        };
    }

    E[] array;
    int head;
    int tail;

    @SuppressWarnings("unchecked")
    public ArrayDeque1(int capacity) {
        array = (E[]) new Object[capacity + 1];
    }

    static int inc(int i, int length) {
        if (i + 1 >= length) {
            return 0;
        }
        return i + 1;
    }

    static int dec(int i, int length) {
        if (i - 1 < 0) {
            return length - 1;
        }
        return i - 1;
    }
}
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数组实现中，如果存储的是基本类型，那么无需考虑内存释放，例如

image-20230110084245095

但如果存储的是引用类型，应当设置该位置的引用为 null，以便内存及时释放

image-20230110084632543

# 二.题目练习

# 1.二叉树的锯齿形层序遍历-力扣 103 题

给你二叉树的根节点 root ，返回其节点值的 锯齿形层序遍历 。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。

image-20230908093831514

输入：root = [3,9,20,null,null,15,7]
输出：[[3],[20,9],[15,7]]
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题解:

public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
    List<List<Integer>> result = new ArrayList<>();
    if (root == null) {
        return result;
    }
    LinkedListQueue<TreeNode> queue = new LinkedListQueue<>();
    queue.offer(root);
    int c1 = 1; // 当前层节点数
    boolean odd = true; // 奇数层
    while (!queue.isEmpty()) {
        LinkedList<Integer> level = new LinkedList<>(); // 保存每一层结果
        int c2 = 0; // 下一层节点数
        for (int i = 0; i < c1; i++) {
            TreeNode n = queue.poll();
            if (odd) {
                level.offerLast(n.val);
            } else {
                level.offerFirst(n.val);
            }
            if (n.left != null) {
                queue.offer(n.left);
                c2++;
            }
            if (n.right != null) {
                queue.offer(n.right);
                c2++;
            }
        }
        odd = !odd;
        result.add(level);
        c1 = c2;
    }
    return result;
}
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# 2.设计双端队列-力扣 641 题

public class LinkedListDeque<E> implements Deque<E>, Iterable<E> {

    @Override
    public boolean offerFirst(E e) {
        if (isFull()) {
            return false;
        }
        Node<E> a = sentinel;
        Node<E> b = sentinel.next;
        Node<E> added = new Node<>(a, e, b);
        a.next = added;
        b.prev = added;
        size++;
        return true;
    }

    @Override
    public boolean offerLast(E e) {
        if (isFull()) {
            return false;
        }
        Node<E> a = sentinel.prev;
        Node<E> b = sentinel;
        Node<E> added = new Node<>(a, e, b);
        a.next = added;
        b.prev = added;
        size++;
        return false;
    }

    // a  b
    @Override
    public E pollFirst() {
        if (isEmpty()) {
            return null;
        }
        Node<E> a = sentinel;
        Node<E> removed = sentinel.next;
        Node<E> b = removed.next;
        a.next = b;
        b.prev = a;
        size--;
        return removed.value;
    }

    // a  b
    @Override
    public E pollLast() {
        if (isEmpty()) {
            return null;
        }
        Node<E> b = sentinel;
        Node<E> removed = sentinel.prev;
        Node<E> a = removed.prev;
        a.next = b;
        b.prev = a;
        size--;
        return removed.value;
    }

    @Override
    public E peekFirst() {
        if (isEmpty()) {
            return null;
        }
        return sentinel.next.value;
    }

    @Override
    public E peekLast() {
        if (isEmpty()) {
            return null;
        }
        return sentinel.prev.value;
    }

    @Override
    public boolean isEmpty() {
        return size == 0;
    }

    @Override
    public boolean isFull() {
        return size == capacity;
    }

    @Override
    public Iterator<E> iterator() {
        return new Iterator<E>() {
            Node<E> p = sentinel.next;

            @Override
            public boolean hasNext() {
                return p != sentinel;
            }

            @Override
            public E next() {
                E value = p.value;
                p = p.next;
                return value;
            }
        };
    }

    static class Node<E> {
        Node<E> prev;
        E value;
        Node<E> next;

        public Node(Node<E> prev, E value, Node<E> next) {
            this.prev = prev;
            this.value = value;
            this.next = next;
        }
    }

    int capacity;
    int size;
    Node<E> sentinel = new Node<>(null, null, null);

    public LinkedListDeque(int capacity) {
        this.capacity = capacity;
        sentinel.next = sentinel;
        sentinel.prev = sentinel;
    }
}
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上次更新: 10/29/2024, 10:27:50 AM

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