232. 用栈实现队列 (Python 实现)

题目:

请你仅使用两个栈实现先入先出队列。队列应当支持一般队列支持的所有操作(push、pop、peek、empty):
实现 MyQueue 类:
void push(int x) 将元素 x 推到队列的末尾
int pop() 从队列的开头移除并返回元素
int peek() 返回队列开头的元素
boolean empty() 如果队列为空,返回 true ;否则,返回 false
说明:
你 只能 使用标准的栈操作 —— 也就是只有 push to top, peek/pop from top, size, 和 is empty 操作是合法的。
你所使用的语言也许不支持栈。你可以使用 list 或者 deque(双端队列)来模拟一个栈,只要是标准的栈操作即可。

示例 1:

输入:
[“MyQueue”, “push”, “push”, “peek”, “pop”, “empty”]
[[], [1], [2], [], [], []]
输出:
[null, null, null, 1, 1, false]
解释:
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false

代码:

class MyQueue:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.stack1 = []
        self.stack2 = []

    def push(self, x: int) -> None:
        """
        Push element x to the back of queue.
        """
        self.stack1.append(x)


    def pop(self) -> int:
        """
        Removes the element from in front of queue and returns that element.
        """
        if not self.stack2:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
        
        return self.stack2.pop()


    def peek(self) -> int:
        """
        Get the front element.
        """
        if not self.stack2:
            while self.stack1:
                self.stack2.append(self.stack1.pop())

        return self.stack2[-1]

    def empty(self) -> bool:
        """
        Returns whether the queue is empty.
        """
        return not self.stack1 and not self.stack2
# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()
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