leetcode1刷题笔记-栈3

用栈实现队列

使用栈实现队列的下列操作：

push(x) -- 将一个元素放入队列的尾部。
pop() -- 从队列首部移除元素。
peek() -- 返回队列首部的元素。
empty() -- 返回队列是否为空。

MyQueue queue = new MyQueue();

queue.push(1);
queue.push(2);  
queue.peek();  // 返回 1
queue.pop();   // 返回 1
queue.empty(); // 返回 false

这道题的解题思路和上一题类似，但是这一题中必须要用两个数组进行模拟，一个数组用于储存push进来的数，一个数组用来储存需要pop出去的数：

class MyQueue:

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


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



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

        """
        
        while not self.stackpop:
            while self.stackpush:
                a=self.stackpush.pop()
                self.stackpop.append(a)
        return self.stackpop.pop()
        
        


    def peek(self) -> int:
        """
        Get the front element.
        """

        while not self.stackpop:
            while self.stackpush:
                a=self.stackpush.pop()
                
                self.stackpop.append(a)
        return self.stackpop[-1]


    def empty(self) -> bool:
        """
        Returns whether the queue is empty.
        """
        if not self.stackpush and not self.stackpop:
            return True
        else:
            return False



# 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()