LeetCode #292 NimGame. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접
술게임 베스킨라빈스 할때 생각했던거랑 똑같음
It's almost same with drinking game "Baskin robins 31" in Korea
LeetCode #292
Q.
You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones.
Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap.
친구와 Nim 게임을 한다. 테이블에 돌무더기가 있는데, 한번에 1~ 3개까지만 제거할 수 있다. 최종 돌을 제거하는 사람이 이긴다. 내 턴이 먼저 시작이다.
둘다 똘똘해서 전략을 짠다. 돌의 숫자를 입력 받았을 때 이길 수 있는지 없는지 확인하는 함수를 짜라.
e.g.
Example:
Input: 4
Output: false
Explanation: If there are 4 stones in the heap, then you will never win the game;
No matter 1, 2, or 3 stones you remove, the last stone will always be
removed by your friend.
설명 : 4를 입력받으면 절대 못이긴다. 1~ 3 어떤 수를 제거하든지, 다음 차례에 친구가 마지막 돌을 제거하기 때문이다.
Process
// Process
//1. Input integer n
//2. Check if it's multiply of 4 or not
// 2.1. If so -> false
// 2.2. If not -> true
//3. Finish
// 처리과정
//1. 돌 개수 정수 n 입력받는다.
//2. 4의 배수인지 확인한다.
// 2.1. 맞으면 -> false
// 2.2. 아니면 -> true
//3. 끝낸다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
bool canWinNim(int n) {
if (n % 4 == 0)
return false;
else
return true;
}
};
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