leetcode 232. Implement Queue using Stacks

题目内容
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:

1. 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.2. 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.3. You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

题目分析

通过栈实现队列的操作。
在这里通过栈列实现队列的操作。队列的特点：先进先出。 栈的特点，后进先出。
通过两个栈来实现，栈s1和栈s2。
在s1中将所有的元素push进去，然后pop出来存入s2中，即可实现队列操作。
代码如下

class MyQueue {    Stack<Integer> s1 = new Stack<>();    Stack<Integer> s2 = new Stack<>();    // 队列的push操作    public void push(int x) {        s1.push(x);    }    // 队列的poo操作，检测s2是否为空，不为空则直接pop，为空则讲s1中的元素push进入s2    public void pop() {        if(!s2.isEmpty())            s2.pop();        else        {            while(!s1.isEmpty())            {                s2.push(s1.pop());            }            s2.pop();        }                   }    // 队列最前元素    public int peek() {        if(!s2.isEmpty())            return s2.peek();        else        {            while(!s1.isEmpty())            {                s2.push(s1.pop());                            }                    }        return s2.peek();    }    // 判断是否为空    public boolean empty() {        if(s1.isEmpty()&&s2.isEmpty())        return true;        return false;    }}