[Leetcode]225. Implement Stack using Queues

225. Implement Stack using Queues

Implement the following operations of a stack using queues.

push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
empty() -- Return whether the stack is empty.
Example:

MyStack stack = new MyStack();

stack.push(1);
stack.push(2);
stack.top(); // returns 2
stack.pop(); // returns 2
stack.empty(); // returns false
Notes:

You must use only standard operations of a queue -- which means only push to back, peek/pop from front, size, and is empty operations are valid.
Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

• 本题难度: Easy
• Topic: Data Structure

Description

我的代码

import collections


class MyStack:
    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.queue1 = collections.deque([])
        self.queue2 = collections.deque([])

    def push(self, x: 'int') -> 'None':
        """
        Push element x onto stack.
        """
        while(not self.empty(self.queue1)):
            self.queue2.append(self.queue1.popleft())
        self.queue1.append(x)
        while(not self.empty(self.queue2)):
            self.queue1.append(self.queue2.popleft())
        return


    def pop(self) -> 'int':
        """
        Removes the element on top of the stack and returns that element.
        """
        return self.queue1.popleft()

    def top(self) -> 'int':
        """
        Get the top element.
        """
        return self.queue1[0]

    def empty(self) -> 'bool':
        """
        Returns whether the stack is empty.
        """
        return self.queue1 == []



        # Your MyStack object will be instantiated and called as such:
        # obj = MyStack()
        # obj.push(x)
        # param_2 = obj.pop()
        # param_3 = obj.top()

思路

push 1
q1 [1]
q2 []

push 2
q1 [2]   -> q1 [2, 1]
q2 [1]   -> q2 []

push 3
q1 [3]   -> q1 [3, 2, 1]
q2 [2,1]   -> q2 []

• 时间复杂度 push O(n) others O(1)