225. Implement Stack using Queues [easy] (Python)

题目链接

https://leetcode.com/problems/implement-stack-using-queues/

题目原文

Implement the following operations of a stack using queues.

1. push(x) – Push element x onto stack.
2. pop() – Removes the element on top of the stack.
3. top() – Get the top element.
4. empty() – Return whether the stack is empty.

Notes:

1. 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.
2. 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.
3. You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

题目翻译

太长不翻了。大概就是用队列的push to back, peek/pop from front, size, is empty方法，实现栈的push, pop, top, empty方法。

思路方法

镜像问题：232. Implement Queue using Stacks

思路一

队列是先进先出，每次出只能出队列的头部，而栈是后进先出，所以可以想办法每次把入队的元素弄到队列头部。于是可以考虑在每次push到队列的时候对其他元素做个重新pop和push将当前元素转移到队头。
该方法需要一个队列，push的复杂度O(n)，pop的复杂度O(1)。

代码

class Stack(object):
    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []

    def push(self, x):
        """
        :type x: int
        :rtype: nothing
        """
        self.stack.insert(0, x)
        for i in range(len(self.stack)-1):
            self.stack.insert(0, self.stack[-1])
            self.stack.pop()

    def pop(self):
        """
        :rtype: nothing
        """
        self.stack.pop()

    def top(self):
        """
        :rtype: int
        """
        return self.stack[-1]

    def empty(self):
        """
        :rtype: bool
        """
        return not self.stack

思路二

也可以用两个队列实现，将思路一的过程分开。但这时要引入一个变量来记录栈顶元素。
该方法需要两个队列，push的复杂度O(1)，pop的复杂度O(n)。

代码

class Stack(object):
    def __init__(self):
        """
        initialize your data structure here.
        """
        self.queue1 = []
        self.queue2 = []
        self.topx = None

    def push(self, x):
        """
        :type x: int
        :rtype: nothing
        """
        self.queue1.insert(0, x)
        self.topx = x

    def pop(self):
        """
        :rtype: nothing
        """
        while len(self.queue1) > 1:
            self.topx = self.queue1.pop()
            self.queue2.insert(0, self.topx)
        self.queue1.pop()
        self.queue1, self.queue2 = self.queue2, self.queue1

    def top(self):
        """
        :rtype: int
        """
        return self.topx

    def empty(self):
        """
        :rtype: bool
        """
        return not self.queue1

思路三

用两个队列，push的复杂度O(n)，pop的复杂度O(1)。

代码

class Stack(object):
    def __init__(self):
        """
        initialize your data structure here.
        """
        self.queue1 = []
        self.queue2 = []

    def push(self, x):
        """
        :type x: int
        :rtype: nothing
        """
        self.queue2.insert(0, x)
        while self.queue1:
            self.queue2.insert(0, self.queue1.pop())
        self.queue1, self.queue2 = self.queue2, self.queue1

    def pop(self):
        """
        :rtype: nothing
        """
        self.queue1.pop()

    def top(self):
        """
        :rtype: int
        """
        return self.queue1[-1]

    def empty(self):
        """
        :rtype: bool
        """
        return not self.queue1

PS: 新手刷LeetCode，新手写博客，写错了或者写的不清楚还请帮忙指出，谢谢！
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