[Leetcode] 225. Implement Stack using Queues

큐를 이용한 스택 구현

225. Implement Stack Using Queues

https://leetcode.com/problems/implement-stack-using-queues/

Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (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 that 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.

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풀이. push()할 때, 큐를 이용해 재정렬

class MyStack:
    def __init__(self):
        self.q = collections.deque()

    def push(self, x):
        self.q.append(x)
        #요소 삽입 후 맨 앞에 두는 상태로 재정렬
        for _ in range(len(self.q) -1):
            self.q.append(self.q.popleft())

    def pop(self):
        return self.q.popleft()

    def top(self):
        return self.q[0]

    def empty(self):
        return len(self.q) == 0