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).

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使用栈来实现一个队列。这个题目之前在《剑指offer》上面见过，没有什么好说的。使用两个栈，一个栈用来保存插入的元素。另外一个栈用来运行pop或top操作，每当运行pop或top操作时检查另外一个栈是否为空，假设为空，将第一栈中的元素所有弹出并插入到第二个栈中。再将第二个栈中的元素弹出就可以。须要注意的是这道题的编程时有一个技巧，能够使用peek来实现pop。这样能够降低反复代码。编程时要能扩展思维，假设因为pop函数的声明再前面就陷入用pop来实现peek的功能的话就会感觉无从下手了。

runtime:0ms

class Queue {

public:

    // Push element x to the back of queue.

    void push(int x) {

        pushStack.push(x);

    }

    // Removes the element from in front of queue.

    void pop(void) {

        peek();//这里能够使用peek进行两个栈之间元素的转移从而避免反复代码

        popStack.pop();

    }

    // Get the front element.

    int peek(void) {

        if(popStack.empty())

        {

            while(!pushStack.empty())

            {

                popStack.push(pushStack.top());

                pushStack.pop();

            }

        }

        return popStack.top();

    }

    // Return whether the queue is empty.

    bool empty(void) {

        return pushStack.empty()&&popStack.empty();

    }

    private:

    stack<int> pushStack;//数据被插入到这个栈中

    stack<int> popStack;//数据从这个栈中弹出

};