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English Version

题目描述

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(pushpoppeekempty):

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 toppeek/pop from topsize, 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。

 

进阶:

  • 你能否实现每个操作均摊时间复杂度为 O(1) 的队列?换句话说,执行 n 个操作的总时间复杂度为 O(n) ,即使其中一个操作可能花费较长时间。

 

示例:

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

 

提示:

  • 1 <= x <= 9
  • 最多调用 100pushpoppeekempty
  • 假设所有操作都是有效的 (例如,一个空的队列不会调用 pop 或者 peek 操作)

解法

Python3

class MyQueue:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.s1 = []
        self.s2 = []


    def push(self, x: int) -> None:
        """
        Push element x to the back of queue.
        """
        self.s1.append(x)


    def pop(self) -> int:
        """
        Removes the element from in front of queue and returns that element.
        """
        self._move()
        return self.s2.pop()

    def peek(self) -> int:
        """
        Get the front element.
        """
        self._move()
        return self.s2[-1]


    def empty(self) -> bool:
        """
        Returns whether the queue is empty.
        """
        return len(self.s1) + len(self.s2) == 0


    def _move(self):
        """
        Move elements from s1 to s2.
        """
        if len(self.s2) == 0:
            while len(self.s1) > 0:
                self.s2.append(self.s1.pop())


# 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()

Java

class MyQueue {

    private Deque<Integer> s1 = new ArrayDeque<>();
    private Deque<Integer> s2 = new ArrayDeque<>();

    /** Initialize your data structure here. */
    public MyQueue() {

    }

    /** Push element x to the back of queue. */
    public void push(int x) {
        s1.push(x);
    }

    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        move();
        return s2.pop();
    }

    /** Get the front element. */
    public int peek() {
        move();
        return s2.peek();
    }

    /** Returns whether the queue is empty. */
    public boolean empty() {
        return s1.isEmpty() && s2.isEmpty();
    }

    /** Move elements from s1 to s2. */
    private void move() {
        if (s2.isEmpty()) {
            while (!s1.isEmpty()) {
                s2.push(s1.pop());
            }
        }
    }
}

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue obj = new MyQueue();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.peek();
 * boolean param_4 = obj.empty();
 */

...