232. 用栈实现队列

题目来源：232. 用栈实现队列

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。

 

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。

 

/**
 * Initialize your data structure here.
 */
var MyQueue = function() {
    this.inStack = [];
    this.outStack = [];
};

/**
 * Push element x to the back of queue. 
 * @param {number} x
 * @return {void}
 */
MyQueue.prototype.push = function(x) {
    this.inStack.push(x);
};

/**
 * Removes the element from in front of queue and returns that element.
 * @return {number}
 */
MyQueue.prototype.pop = function() {
    if(!this.outStack.length){
        this.in2out();
    }
    return this.outStack.pop();
};

/**
 * Get the front element.
 * @return {number}
 */
MyQueue.prototype.peek = function() {
    if(!this.outStack.length){
        this.in2out();
    }
    return this.outStack[this.outStack.length - 1];
};

/**
 * Returns whether the queue is empty.
 * @return {boolean}
 */
MyQueue.prototype.empty = function() {
    return this.inStack.length == 0 && this.outStack.length == 0;
};

/**
 * instack to outstack
 * @return {void}
 */
MyQueue.prototype.in2out = function() {
    while(this.inStack.length){
        this.outStack.push(this.inStack.pop());
    }
};

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

 

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

 Python3

class MyQueue:
  
    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.inStack = list()
        self.outStack = list()


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


    def in2out(self) -> None:
        """
        instack to outstack.
        """
        while self.inStack:
          self.outStack.append(self.inStack.pop())


    def pop(self) -> int:
        """
        Removes the element from in front of queue and returns that element.
        """
        if not self.outStack:
          self.in2out()
        return self.outStack.pop()


    def peek(self) -> int:
        """
        Get the front element.
        """
        if not self.outStack:
          self.in2out()
        return self.outStack[-1]


    def empty(self) -> bool:
        """
        Returns whether the queue is empty.
        """
        return not self.inStack and not self.outStack



# Your MyQueue object will be instantiated and called as such:
if __name__ == '__main__':
  obj = MyQueue()
  x = 1
  obj.push(x)
  obj.push(3)
  obj.push(2)
  print(obj.inStack, obj.outStack)
  param_2 = obj.pop()
  print(obj.inStack, obj.outStack, param_2)
  param_3 = obj.peek()
  print(obj.inStack, obj.outStack, param_3)
  param_4 = obj.empty()
  print(obj.inStack, obj.outStack, param_4)

 

提示：

• 1 <= x <= 9
• 最多调用 100 次 push、pop、peek 和 empty
• 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）