2021-07-06用队列实现栈

leetcode每日一题之用队列实现栈

题目链接：https://leetcode-cn.com/problems/implement-stack-using-queues/

题目描述：

请你仅使用两个队列实现一个后入先出（LIFO）的栈，并支持普通栈的全部四种操作（push、top、pop 和 empty）。

实现 MyStack 类：

void push(int x) 将元素 x 压入栈顶。
int pop() 移除并返回栈顶元素。
int top() 返回栈顶元素。
boolean empty() 如果栈是空的，返回 true ；否则，返回 false 。

注意：

你只能使用队列的基本操作 —— 也就是 push to back、peek/pop from front、size 和 is empty 这些操作。
你所使用的语言也许不支持队列。 你可以使用 list （列表）或者 deque（双端队列）来模拟一个队列 , 只要是标准的队列操作即可。

示例：

输入：
["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]
输出：
[null, null, null, 2, 2, false]
解释：
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // 返回 2
myStack.pop(); // 返回 2
myStack.empty(); // 返回 False

思路：队列是只允许从队列尾部插入，队列头部出的数据结构，并且是先进先出

栈是先进后出，从栈顶插入并且弹出的数据结构

如下图：

代码如下：

    package com.tao.queue;
    import java.util.LinkedList;
    import java.util.Queue;
    /**
     * @Classname MyStack
     * @Description 用队列实现栈
     * @Date 2021/7/6 10:54
     * @Author Anonymous
     */
    @SuppressWarnings("all")
    /**
     * Your MyStack object will be instantiated and called as such:
     * MyStack obj = new MyStack();
     * obj.push(x);
     * int param_2 = obj.pop();
     * int param_3 = obj.top();
     * boolean param_4 = obj.empty();
     */
    public class MyStack {
        Queue<Integer> queue;
    
        /**
         * Initialize your data structure here.
         */
        public MyStack() {
            queue = new LinkedList<Integer>();
        }
    
        /**
         * Push element x onto stack.
         */
        public void push(int x) {
            //相当于每次放元素都放到队列头部，因为队列是从队列尾部放入元素，队列头部出来，先进先出
            //而栈是放入栈顶，并且从栈顶出元素，先进后出
            queue.offer(x);
            for (int i = 1; i < queue.size(); i++) {
                queue.offer(queue.poll());
            }
        }
    
        /**
         * Removes the element on top of the stack and returns that element.
         */
        public int pop() {
            return queue.poll();
        }
    
        /**
         * Get the top element.
         */
        public int top() {
            return queue.peek();
        }
    
        /**
         * Returns whether the stack is empty.
         */
        public boolean empty() {
            return queue.isEmpty();
        }
    }

栈的源码如下：

    /*
     * Copyright (c) 1994, 2010, Oracle and/or its affiliates. All rights reserved.
     * ORACLE PROPRIETARY/CONFIDENTIAL. Use is subject to license terms.
     */
    package java.util;
    /**
     * The <code>Stack</code> class represents a last-in-first-out
     * (LIFO) stack of objects. It extends class <tt>Vector</tt> with five
     * operations that allow a vector to be treated as a stack. The usual
     * <tt>push</tt> and <tt>pop</tt> operations are provided, as well as a
     * method to <tt>peek</tt> at the top item on the stack, a method to test
     * for whether the stack is <tt>empty</tt>, and a method to <tt>search</tt>
     * the stack for an item and discover how far it is from the top.
     * <p>
     * When a stack is first created, it contains no items.
     *
     * <p>A more complete and consistent set of LIFO stack operations is
     * provided by the {@link Deque} interface and its implementations, which
     * should be used in preference to this class.  For example:
     * <pre>   {@code
     *   Deque<Integer> stack = new ArrayDeque<Integer>();}</pre>
     *
     * @author  Jonathan Payne
     * @since   JDK1.0
     */
    public
    class Stack<E> extends Vector<E> {
        /**
         * Creates an empty Stack.
         */
        public Stack() {
        }
    
        /**
         * Pushes an item onto the top of this stack. This has exactly
         * the same effect as:
         * <blockquote><pre>
         * addElement(item)</pre></blockquote>
         *
         * @param   item   the item to be pushed onto this stack.
         * @return  the <code>item</code> argument.
         * @see     java.util.Vector#addElement
         */
        public E push(E item) {
            addElement(item);
    
            return item;
        }
    
        /**
         * Removes the object at the top of this stack and returns that
         * object as the value of this function.
         *
         * @return  The object at the top of this stack (the last item
         *          of the <tt>Vector</tt> object).
         * @throws  EmptyStackException  if this stack is empty.
         */
        public synchronized E pop() {
            E       obj;
            int     len = size();
    
            obj = peek();
            removeElementAt(len - 1);
    
            return obj;
        }
    
        /**
         * Looks at the object at the top of this stack without removing it
         * from the stack.
         *
         * @return  the object at the top of this stack (the last item
         *          of the <tt>Vector</tt> object).
         * @throws  EmptyStackException  if this stack is empty.
         */
        public synchronized E peek() {
            int     len = size();
    
            if (len == 0)
                throw new EmptyStackException();
            return elementAt(len - 1);
        }
    
        /**
         * Tests if this stack is empty.
         *
         * @return  <code>true</code> if and only if this stack contains
         *          no items; <code>false</code> otherwise.
         */
        public boolean empty() {
            return size() == 0;
        }
    
        /**
         * Returns the 1-based position where an object is on this stack.
         * If the object <tt>o</tt> occurs as an item in this stack, this
         * method returns the distance from the top of the stack of the
         * occurrence nearest the top of the stack; the topmost item on the
         * stack is considered to be at distance <tt>1</tt>. The <tt>equals</tt>
         * method is used to compare <tt>o</tt> to the
         * items in this stack.
         *
         * @param   o   the desired object.
         * @return  the 1-based position from the top of the stack where
         *          the object is located; the return value <code>-1</code>
         *          indicates that the object is not on the stack.
         */
        public synchronized int search(Object o) {
            int i = lastIndexOf(o);
    
            if (i >= 0) {
                return size() - i;
            }
            return -1;
        }
    
        /** use serialVersionUID from JDK 1.0.2 for interoperability */
        private static final long serialVersionUID = 1224463164541339165L;
    }

解法2：用两个队列实现

链接：https://leetcode-cn.com/problems/implement-stack-using-queues/solution/wu-tu-guan-fang-tui-jian-ti-jie-yong-dui-63d4/

代码：

    class MyStack {
        Queue<Integer> queue1;
        Queue<Integer> queue2;
    
        public MyStack() {
            queue1 = new LinkedList<Integer>();
            queue2 = new LinkedList<Integer>();
        }
        
        public void push(int x) {
            queue2.offer(x);//先将元素假如到辅助队列中
            while (!queue1.isEmpty()) {
                queue2.offer(queue1.poll());
            }
            //下面是交换队列和辅助队列
            Queue<Integer> temp = queue1;
            queue1 = queue2;
            queue2 = temp;
        }
        
        public int pop() {
            return queue1.poll();
        }
        
        public int top() {
            return queue1.peek();
        }
        
        public boolean empty() {
            return queue1.isEmpty();
        }
    }