개요

LeetCode #232, Implement Queue Using Stacks 문제를 풀어봅니다.

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.

pop(), peek()은 결국 Queue에서 건드리는 위치가 같습니다. pop()을 할 때 peek()를 호출하고, peek()에서는 output이 없을 경우 input에 있는 것들을 output에 모두 재입력하도록 합니다.

class MyQueue:

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

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

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

    def peek(self) -> int:
        """
        Get the front element.
        """
        # output 없으면 모두 재입력
        if not self.output:
            while self.input:
                self.output.append(self.input.pop())
        return self.output[-1]

    def empty(self) -> bool:
        """
        Returns whether the queue is empty.
        """
        return self.input == [] and self.output == []