programmers_3 x n 타일링_java

https://school.programmers.co.kr/learn/courses/30/lessons/12902

프로그래머스

class Solution {
    
    private final int mod = 1000000007;
    
    public int solution(int n) {
        int answer = 0;
        
        long[] dp = new long[n+1];
        
        // f(n)= f(n-2)*4 - f(n-4)
        dp[0]=1;
        dp[2]=3;
        
        for(int i=4; i<=n; i+=2){
            dp[i]= (dp[i-2]*4%mod -dp[i-4]%mod +mod)%mod;
        };
        
        answer = (int)dp[n];
        
        return answer;
    }
}