Leetcodepython3简单问题225。使用队列的实现堆栈,leetcodepython3,225ImplementStackusingQueues


1.编辑器

我使用的是win10+vscode+leetcode+python3
环境配置参见我的博客:
链接

2.第二百二十五题

(1)题目
英文:

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

中文:
使用队列实现栈的下列操作:

push(x) – 元素 x 入栈
pop() – 移除栈顶元素
top() – 获取栈顶元素
empty() – 返回栈是否为空
注意:

你只能使用队列的基本操作-- 也就是 push to back, peek/pop from front, size, 和 is empty 这些操作是合法的。
你所使用的语言也许不支持队列。 你可以使用 list 或者 deque(双端队列)来模拟一个队列 , 只要是标准的队列操作即可。
你可以假设所有操作都是有效的(例如, 对一个空的栈不会调用 pop 或者 top 操作)。

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/implement-stack-using-queues

(2)解法
① 单个list完成,每次都要把最新push进去的元素放在首位,也就是栈首
(耗时:36ms,内存:13.5M)

class MyStack:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.q = []

    def push(self, x: int) -> None:
        """
        Push element x onto stack.
        """
        self.q.append(x)
        q_length = len(self.q)
        while q_length > 1:
            self.q.append(self.q.pop(0))
            q_length -= 1

    def pop(self) -> int:
        """
        Removes the element on top of the stack and returns that element.
        """
        return self.q.pop(0)

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


    def empty(self) -> bool:
        """
        Returns whether the stack is empty.
        """
        return not bool(self.q)





# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()

② deque
(耗时:36ms,内存:13.5M)

from collections import deque
class MyStack:
    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.data = deque()
        self.help = deque()
    def push(self, x: int) -> None:
        """
        Push element x onto stack.
        """
        self.data.append(x)
    def pop(self) -> int:
        """
        Removes the element on top of the stack and returns that element.
        """
        while len(self.data) > 1:
            self.help.append(self.data.popleft())
        tmp = self.data.popleft()        
        self.help,self.data = self.data,self.help
        return tmp
    def top(self) -> int:
        """
        Get the top element.
        """
        while len(self.data) != 1:
            self.help.append(self.data.popleft())
        tmp = self.data.popleft()
        self.help.append(tmp)
        self.help,self.data = self.data,self.help
        return tmp
    def empty(self) -> bool:
        """
        Returns whether the stack is empty.
        """
        return not bool(self.data)

# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()

注意:
1. popleft() 只能给 deque用,deque建立的是deque([]),并使用append进行元素的添加。
2.top函数可以改为更简单的:

    def top(self) -> int:
        """
        Get the top element.
        """
        return self.data[-1]

3.deque中没有 popright 只有 pop