leetcode-232.Implement-Queue-using-Stacks

一、题目

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).

Implement the MyQueue class:

• void push(int x) Pushes element x to the back of the queue.
• int pop() Removes the element from the front of the queue and returns it.
• int peek() Returns the element at the front of the queue.
• boolean empty() Returns true if the queue is empty, false otherwise.

Notes:

• You must use only standard operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid.
• Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack’s standard operations.

Follow-up: Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.

Example 1:

Input
[“MyQueue”, “push”, “push”, “peek”, “pop”, “empty”]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
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

Constraints:

• 1 <= x <= 9
• At most 100 calls will be made to push, pop, peek, and empty.
• All the calls to pop and peek are valid.

二、思路

参考了大佬的这篇题解

栈是一种LIFO的数据结构，而队列是FIFO的数据结构，我们用栈实现队列就是LIFO => FIFO，我们可以用两个栈来实现。

首先我们定义一个栈stack1用于存放原始的数据，即接收push()方法操作的数据；

接着我们定义另外一个栈stack2，并将stack1中的数据使用pop()方法传送到stack2中，这样这个stack2就类似于我们的队列了：

• 对于stack1来说，stack1的栈顶元素第一个弹栈，stack1的栈底元素是最后一个弹栈的；
• 对于stack2来说，接收到的第一个元素（stack1的栈顶元素）作为栈底元素，接收到的最后一个元素（stack的栈底元素）作为栈顶元素；
• 这样，当我们对stack2进行pop()操作时，返回的数是我们第一个传入的数，即队列的第一个元素。

三、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()