225. Implement Stack using Queues (Python)

Related Topic

Stack.

Description

Implement the following operations of a stack using queues.

• push(x) – Push element x onto stack.
• pop() – Removes the element on top of the stack.
• top() – Get the top element.
• empty() – Return whether the stack is empty.

Sample I/O

Example 1

MyStack stack = new MyStack();

stack.push(1);
stack.push(2);  
stack.top();   // returns 2
stack.pop();   // returns 2
stack.empty(); // returns false

Note

1. You must use only standard operations of a queue – which means only push to back, top/pop from front, size, and is empty operations are valid.
2. Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
3. You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

Methodology

Queue is FIFO. We can use python deque to act as queue. The key point is when you append new element to the deque. we need to move all x’s previous elements to x’s next. So we use for loop to control the moving length and popleft each element one by one and append to the original queue.

class MyStack:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self._queue = collections.deque()

    def push(self, x: int) -> None:
        """
        Push element x onto stack.
        """
        self._queue.append(x)
        for _ in range(len(self._queue)-1):
            self._queue.append(self._queue.popleft())
        

    def pop(self) -> int:
        """
        Removes the element on top of the stack and returns that element.
        """
        return self._queue.popleft()
        

    def top(self) -> int:
        """
        Get the top element.
        """
        return self._queue[0]
        

    def empty(self) -> bool:
        """
        Returns whether the stack is empty.
        """
        return len(self._queue)==0 

BigO

This will give O(n) time complexity.