问题描述

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).

问题分析

这是一道Easy题，但是我并没有看懂题意……也是有一种每天蠢出新境界的感觉。参考：csdn-哈哈的个人专栏
思路是准备a、b两个栈，a负责入栈，b负责出栈。

• push(x) -- 将x加到a栈的末尾；
• pop() -- 判断b栈是否为空，若不为空，则弹出栈顶元素，若为空，则将a栈中的元素依次弹出并加入b栈，再弹出b栈的栈顶元素；
• peek() -- 判断b栈是否为空，若不为空，则返回栈顶元素，若为空，则将a栈中的元素依次弹出并加入b栈，再返回b栈的栈顶元素；
• empty() -- a栈b栈均空时表明队列为空。

AC代码

class Queue(object):
    def __init__(self):
        """
        initialize your data structure here.
        """
        self.a = []
        self.b = [] 

    def push(self, x):
        """
        :type x: int
        :rtype: nothing
        """
        self.a.append(x)
        
    def pop(self):
        """
        :rtype: nothing
        """
        if len(self.b) != 0: 
            return self.b.pop()
        else:
            while len(self.a) != 0:
                self.b.append(self.a.pop())
            return self.b.pop()  

    def peek(self):
        """
        :rtype: int
        """
        if len(self.b) != 0: 
            return self.b[-1]
        else:
            while len(self.a) != 0:
                self.b.append(self.a.pop())
            return self.b[-1]

    def empty(self):
        """
        :rtype: bool
        """
        return len(self.a) == 0 and len(self.b) == 0

Runtime: 44 ms, which beats 71.08% of Python submissions.