790. 多米诺和托米诺平铺

Posted 2022-08-22 lasclocker

public int numTilings(int N)  // dp转移方程: dp(n) = 2 * dp(n-1) + dp(n-3), 时间复杂度O(N), 空间复杂度O(1) 
        if (N == 0) 
            return 0;
        
        if (N == 1) 
            return 1;
        
        if (N == 2) 
            return 2;
        
        int mod = (int)(Math.pow(10, 9) + 7);
        int n1 = 1, n2 = 2, n3 = 5;
        for (int i = 4; i <= N; i++) 
            int temp = (2 * n3 % mod + n1) % mod; // 2*n3+n1 会超过int的最大值(2147483647, 2开头10位数), 所以要对 2*n3 进行mod(1000000007, 1开头10位数)
            n1 = n2;
            n2 = n3;
            n3 = temp;
        
        return n3;