假设我们有两个堆栈,没有其他临时变量。

是否有可能“构造”一个队列数据结构只使用两个堆栈?


当前回答

对于c#开发人员,这里是完整的程序:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace QueueImplimentationUsingStack
{
    class Program
    {
        public class Stack<T>
        {
            public int size;
            public Node<T> head;
            public void Push(T data)
            {
                Node<T> node = new Node<T>();
                node.data = data;
                if (head == null)
                    head = node;
                else
                {
                    node.link = head;
                    head = node;
                }
                size++;
                Display();
            }
            public Node<T> Pop()
            {
                if (head == null)
                    return null;
                else
                {
                    Node<T> temp = head;
                    //temp.link = null;
                    head = head.link;
                    size--;
                    Display();
                    return temp;
                }
            }
            public void Display()
            {
                if (size == 0)
                    Console.WriteLine("Empty");
                else
                {
                    Console.Clear();
                    Node<T> temp = head;
                    while (temp!= null)
                    {
                        Console.WriteLine(temp.data);
                        temp = temp.link;
                    }
                }
            }
        }

        public class Queue<T>
        {
            public int size;
            public Stack<T> inbox;
            public Stack<T> outbox;
            public Queue()
            {
                inbox = new Stack<T>();
                outbox = new Stack<T>();
            }
            public void EnQueue(T data)
            {
                inbox.Push(data);
                size++;
            }
            public Node<T> DeQueue()
            {
                if (outbox.size == 0)
                {
                    while (inbox.size != 0)
                    {
                        outbox.Push(inbox.Pop().data);
                    }
                }
                Node<T> temp = new Node<T>();
                if (outbox.size != 0)
                {
                    temp = outbox.Pop();
                    size--;
                }
                return temp;
            }

        }
        public class Node<T>
        {
            public T data;
            public Node<T> link;
        }

        static void Main(string[] args)
        {
            Queue<int> q = new Queue<int>();
            for (int i = 1; i <= 3; i++)
                q.EnQueue(i);
           // q.Display();
            for (int i = 1; i < 3; i++)
                q.DeQueue();
            //q.Display();
            Console.ReadKey();
        }
    }
}

其他回答

// Two stacks s1 Original and s2 as Temp one
    private Stack<Integer> s1 = new Stack<Integer>();
    private Stack<Integer> s2 = new Stack<Integer>();

    /*
     * Here we insert the data into the stack and if data all ready exist on
     * stack than we copy the entire stack s1 to s2 recursively and push the new
     * element data onto s1 and than again recursively call the s2 to pop on s1.
     * 
     * Note here we can use either way ie We can keep pushing on s1 and than
     * while popping we can remove the first element from s2 by copying
     * recursively the data and removing the first index element.
     */
    public void insert( int data )
    {
        if( s1.size() == 0 )
        {
            s1.push( data );
        }
        else
        {
            while( !s1.isEmpty() )
            {
                s2.push( s1.pop() );
            }
            s1.push( data );
            while( !s2.isEmpty() )
            {
                s1.push( s2.pop() );
            }
        }
    }

    public void remove()
    {
        if( s1.isEmpty() )
        {
            System.out.println( "Empty" );
        }
        else
        {
            s1.pop();

        }
    }

使用堆栈实现队列的以下操作。

push(x)——将元素x推到队列的后面。

pop()——从队列前面移除元素。

peek()——获取前端元素。

empty()——返回队列是否为空。

class MyQueue {

  Stack<Integer> input;
  Stack<Integer> output;

  /** Initialize your data structure here. */
  public MyQueue() {
    input = new Stack<Integer>();
    output = new Stack<Integer>();
  }

  /** Push element x to the back of queue. */
  public void push(int x) {
    input.push(x);
  }

  /** Removes the element from in front of queue and returns that element. */
  public int pop() {
    peek();
    return output.pop();
  }

  /** Get the front element. */
  public int peek() {
    if(output.isEmpty()) {
        while(!input.isEmpty()) {
            output.push(input.pop());
        }
    }
    return output.peek();
  }

  /** Returns whether the queue is empty. */
  public boolean empty() {
    return input.isEmpty() && output.isEmpty();
  }
}

简单的JS解决方案**

注:我从其他人的评论中获得了一些想法

/* enQueue(q, x) 1) Push x to stack1 (assuming size of stacks is unlimited). deQueue(q) 1) If both stacks are empty then error. 2) If stack2 is empty While stack1 is not empty, push everything from stack1 to stack2. 3) Pop the element from stack2 and return it. */ class myQueue { constructor() { this.stack1 = []; this.stack2 = []; } push(item) { this.stack1.push(item) } remove() { if (this.stack1.length == 0 && this.stack2.length == 0) { return "Stack are empty" } if (this.stack2.length == 0) { while (this.stack1.length != 0) { this.stack2.push(this.stack1.pop()) } } return this.stack2.pop() } peek() { if (this.stack2.length == 0 && this.stack1.length == 0) { return 'Empty list' } if (this.stack2.length == 0) { while (this.stack1.length != 0) { this.stack2.push(this.stack1.pop()) } } return this.stack2[0] } isEmpty() { return this.stack2.length === 0 && this.stack1.length === 0; } } const q = new myQueue(); q.push(1); q.push(2); q.push(3); q.remove() console.log(q)

public class QueueUsingStacks<T>
{
    private LinkedListStack<T> stack1;
    private LinkedListStack<T> stack2;

    public QueueUsingStacks()
    {
        stack1=new LinkedListStack<T>();
        stack2 = new LinkedListStack<T>();

    }
    public void Copy(LinkedListStack<T> source,LinkedListStack<T> dest )
    {
        while(source.Head!=null)
        {
            dest.Push(source.Head.Data);
            source.Head = source.Head.Next;
        }
    }
    public void Enqueue(T entry)
    {

       stack1.Push(entry);
    }
    public T Dequeue()
    {
        T obj;
        if (stack2 != null)
        {
            Copy(stack1, stack2);
             obj = stack2.Pop();
            Copy(stack2, stack1);
        }
        else
        {
            throw new Exception("Stack is empty");
        }
        return obj;
    }

    public void Display()
    {
        stack1.Display();
    }


}

对于每一个入队列操作,我们都将其添加到stack1的顶部。每次出队列时,我们都将stack1的内容清空到stack2中,并删除堆栈顶部的元素。出队列的时间复杂度是O(n),因为我们必须将stack1复制到stack2。队列的时间复杂度与常规堆栈相同

您甚至可以只使用一个堆栈模拟一个队列。第二个(临时)堆栈可以通过对insert方法的递归调用的调用堆栈来模拟。

在队列中插入新元素时,原理保持不变:

您需要将元素从一个堆栈转移到另一个临时堆栈,以反转它们的顺序。 然后将要插入的新元素推入临时堆栈 然后将元素转移回原始堆栈 新元素将在堆栈的底部,而最老的元素在顶部(第一个被弹出)

一个Queue类只使用一个Stack,如下所示:

public class SimulatedQueue<E> {
    private java.util.Stack<E> stack = new java.util.Stack<E>();

    public void insert(E elem) {
        if (!stack.empty()) {
            E topElem = stack.pop();
            insert(elem);
            stack.push(topElem);
        }
        else
            stack.push(elem);
    }

    public E remove() {
        return stack.pop();
    }
}