LeetCode 232：Implement Queue using Stacks


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).

//题目描述:
//使用栈实现队列的下列操作：
//push(x) --将元素x加至队列尾部
//pop( ) --从队列头部移除元素
//peek( ) --获取队头元素
//empty( ) --返回队列是否为空
//注意：你只可以使用栈的标准操作，这意味着只有push to top（压栈）, peek / pop from top（取栈顶 / 弹栈顶），
//以及empty（判断是否为空）是允许的，取决于你的语言，stack可能没有被内建支持。你可以使用list（列表）或者deque（双端队列）来模拟，
//确保只使用栈的标准操作即可，你可以假设所有的操作都是有效的（例如，不会对一个空的队列执行pop或者peek操作）

//解题方法：用两个栈就可以模拟一个队列，基本思路是两次后进先出 = 先进先出，
//元素入队列总是入in栈，元素出队列如果out栈不为空直接弹出out栈头元素；
//如果out栈为空就把in栈元素出栈全部压入out栈，再弹出out栈头，这样就模拟出了一个队列。
//核心就是保证每个元素出栈时都经过了in，out两个栈，这样就实现了两次后进先出=先进先出。
class Queue {
public:
	stack<int> in;
	stack<int> out;
	void move(){   //将in栈内的所有元素移动到out栈
		while (!in.empty()){
			int x = in.top();
			in.pop();
			out.push(x);
		}
	}

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

	// Removes the element from in front of queue.
	void pop(void) {
		if (out.empty()){
			move();
		}
			if (!out.empty()){
				out.pop();
			}
		}

	// Get the front element.
	int peek(void) {
		if (out.empty()){
			move();
		}
		if (!out.empty()){
			return out.top();
		}
	}

	// Return whether the queue is empty.
	bool empty(void) {
		return in.empty() && out.empty();
	}
};