📌 Problem

Implement a last in first out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal queue (push, top, pop, and empty).

Implement the MyStack class:

• void push(int x) Pushes element x to the top of the stack.
• int pop() Removes the element on the top of the stack and returns it.
• int top() Returns the element on the top of the stack.
• boolean empty() Returns true if the stack is empty, false otherwise.

Notes:

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

Example 1:

Input
["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 2, 2, false]

Explanation
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // return 2
myStack.pop(); // return 2
myStack.empty(); // return False

Constraints:

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

Follow-up: Can you implement the stack 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. You can use more than two queues.

leetcode 에서 풀기

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📝 Solution

from collections import deque 
class MyStack:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.__main_queue=deque()
        self.__sub_queue=deque()
        

    def push(self, x: int) -> None:
        """
        Push element x onto stack.
        """
        self.__main_queue.append(x) # Push to back .
        

    def pop(self) -> int:
        """
        Removes the element on top of the stack and returns that element.
        """
        ret=None
        while len(self.__main_queue)>1:
            temp=self.__main_queue.popleft() # Peek/pop from front . 
            self.__sub_queue.append(temp) # Move main_queue to sub_queue
        
        if len(self.__main_queue) == 1:
            ret=self.__main_queue.popleft()
            self.__main_queue,self.__sub_queue=self.__sub_queue,self.__main_queue # Swap main_queue,sub_queue
        
        return ret
        
    

    def top(self) -> int:
        """
        Get the top element.
        """
        return self.__main_queue[-1]
        

    def empty(self) -> bool:
        """
        Returns whether the stack is empty.
        """
        return False if self.__main_queue else True
        


# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()

• push 는main_queue에 담는다.

  

• pop 은 main_queue 가 하나 남을 때 까지(last-in) sub_queue에 옮겨 주고 pop 의 결과 저장 후 main_queue 와 sub_queue 역할을 변경한다.

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📚 Reference

The figures for header # Solution