[Python] leetcode-1823.Find the Winner of the Circular Game(큐)

문제 설명

There are n friends that are playing a game. The friends are sitting in a circle and are numbered from 1 to n in clockwise order. More formally, moving clockwise from the ith friend brings you to the (i+1)th friend for 1 <= i < n, and moving clockwise from the nth friend brings you to the 1st friend.

The rules of the game are as follows:

1. Start at the 1st friend.
2. Count the next k friends in the clockwise direction including the friend you started at. The counting wraps around the circle and may count some friends more than once.
3. The last friend you counted leaves the circle and loses the game.
4. If there is still more than one friend in the circle, go back to step 2 starting from the friend immediately clockwise of the friend who just lost and repeat.
5. Else, the last friend in the circle wins the game.

Given the number of friends, n, and an integer k, return the winner of the game.

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제한 사항

• 1 <= k <= n <= 500

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입출력 예제

n	k	return
5	2	3
6	5	1

1. 첫번째 예제 이미지
   설명
   1) Start at friend 1.
   2) Count 2 friends clockwise, which are friends 1 and 2.
   3) Friend 2 leaves the circle. Next start is friend 3.
   4) Count 2 friends clockwise, which are friends 3 and 4.
   5) Friend 4 leaves the circle. Next start is friend 5.
   6) Count 2 friends clockwise, which are friends 5 and 1.
   7) Friend 1 leaves the circle. Next start is friend 3.
   8) Count 2 friends clockwise, which are friends 3 and 5.
   9) Friend 5 leaves the circle. Only friend 3 is left, so they are the winner.

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큐를 사용해서 접근했다. 큐에 n명의 친구들을 순서대로 넣어 원소가 하나만 남을 때까지 while문으로 반복했다. k번째가 아닌 친구들이 큐의 첫번째일 경우에는 popleft를 한 후, 다시 append로 뒤로 보냈다. 만약, k번째이면 append는 생략하고 popleft만 해서 큐에서 제거했다.

첫번째 예제의 경우 아래의 이미지와 같은 방식으로 적용된다.

코드 작성

from collections import deque

class Solution:
    def findTheWinner(self, n: int, k: int) -> int:
        q=deque([i for i in range(n)])
        i=0
        while len(q)>1:
            if i!=k-1:
                num=q.popleft()
                q.append(num)
                i+=1
            else:
                q.popleft()
                i=0
        return q[0]+1

결과