【LeetCode OJ 232】Implement Queue using Stacks

题目链接：https://leetcode.com/problems/implement-queue-using-stacks/

题目：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).

解题思路：题目要求用栈来实现队列，并实现其入队列、出队列、查看队首元素和判断队列是否为空四个操作。大致的思路是：用两个栈来模拟实现队列，栈s1作为存储空间，栈s2作为临时缓冲区。

入队时：将元素压入s1

出队时：首先判断s2是否为空，如不为空，则直接弹出栈顶元素，如果为空，则将s1的元素逐个"倒入"s2，把最后一个元素弹出并出队。

示例代码如下：

/** * 用两个栈模拟实现队列基本操作 * @author 徐剑 * @date 2016-02-25 * */public class Solution{Stack<Integer> s1 = new Stack<>();Stack<Integer> s2 = new Stack<>();// Push element x to the back of queue.public void push(int x){s1.push(x);}// Removes the element from in front of queue.public void pop(){if (!s2.empty()){s2.pop();} else if (s1.isEmpty()){return;} else{while (!s1.isEmpty()){int temp = s1.pop();s2.push(temp);}s2.pop();}}// Get the front element.public int peek(){if (!s2.isEmpty())return s2.pop();else if (s1.isEmpty()){return -1;} else{while (!s1.isEmpty()){int temp = s1.pop();s2.push(temp);}return s2.peek();}}// Return whether the queue is empty.public boolean empty(){return s1.isEmpty() && s2.isEmpty();}}