232. Implement Queue using Stacks

最新推荐文章于 2023-10-09 00:28:50 发布

转载最新推荐文章于 2023-10-09 00:28:50 发布 · 37 阅读

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原文链接：http://www.cnblogs.com/zhangfanxmian/p/6929970.html

本文介绍了一种使用两个栈来模拟队列操作的方法。通过将元素先压入第一个栈，再转移到第二个栈的方式，实现了队列的基本操作：push、pop、peek及empty。这种方法巧妙地利用了栈的特性，确保了元素进出的顺序符合队列的要求。

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Implement the following operations of a queue using stacks.

push(x) -- Push element x to the back of queue.
pop() -- Removes the element from in front of queue.
peek() -- Get the front element.
empty() -- Return whether the queue is empty.

Notes:

You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is empty operations are valid.
Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

思路：想要使用堆栈实现队列，堆栈出一个再进一个相当于没进没出，所以我们采用两个堆栈实现队列，先进第一个堆栈，再把第一个堆栈的元素依次给另外一个堆栈，那么即将要进来的元素就到了第一个堆栈的栈底，刚好，再从堆栈2到堆栈1，和队列的顺序就保持一致了。

public class MyQueue { Stack<Integer>stack1=new Stack<>(); Stack<Integer>stack2=new Stack<> /** Initialize your data structure here. */

    public MyQueue() {
        
    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        while(!stack1.isEmpty())
        {
        	stack2.push(stack1.pop());//堆栈1的元素依次给堆栈2
        }
        stack1.push(x);//x就成了1的栈底元素

        while(!stack2.isEmpty())
        {
        	stack1.push(stack2.pop());//再把2给1
        }
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
         return stack1.pop();//栈顶元素就是队列的第一个元素
    }
    
    /** Get the front element. */
    public int peek() {
                return stack1.peek();
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        return stack1.isEmpty();
    }
}

　　堆栈和队列之间的转换和巧妙，大家可以尝试一下！

转载于:https://www.cnblogs.com/zhangfanxmian/p/6929970.html