LeetCode 232. 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).

思路：
用两个栈实现队列，s1枯是队尾，s1的底是队头。在pop时，要借助s2，将s1的数都放到s2上，然后将s2的顶（即s1的底）pop; 取peek时，即将pop时的动作改为top即可；在push时，首先要将s2中的数再还原到s1，然后push;判断是否为empty要s1和s2都为empty时，队列才empty。

代码：

class MyQueue {public:/** Initialize your data structure here. */MyQueue() {}/** Push element x to the back of queue. */void push(int x) {while (!s2.empty()){//如果s2不为空s1.push(s2.top());//那就就先要将s2中的数据全都放到s1中s2.pop();}s1.push(x);//入队列}/** Removes the element from in front of queue and returns that element. */int pop() {while (!s1.empty()){//如果s1不为空s2.push(s1.top());//要将s1中的数据放到s2中，再取s2的栈顶元素s2.top()s1.pop();//再pop}int result = s2.top();//出队列s2.pop();return result;}/** Get the front element. */int peek() {while (!s1.empty()){//如果s1不为空s2.push(s1.top());//要将s1中的数据放到s2中，再取s2的栈顶元素s2.top()s1.pop();//再pop}return s2.top();}/** Returns whether the queue is empty. */bool empty() {return s1.empty()&&s2.empty();//两个栈都为空时，队列才空}
private:stack<int> s1;stack<int> s2;
};/*** Your MyQueue object will be instantiated and called as such:* MyQueue obj = new MyQueue();* obj.push(x);* int param_2 = obj.pop();* int param_3 = obj.peek();* bool param_4 = obj.empty();*/

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