790. Domino and Tromino Tiling

We have two types of tiles: a 2x1 domino shape, and an "L" tromino shape. These shapes may be rotated.

XX  <- domino

XX  <- "L" tromino
X

Given N, how many ways are there to tile a 2 x N board? Return your answer modulo 10^9 + 7.

(In a tiling, every square must be covered by a tile. Two tilings are different if and only if there are two 4-directionally adjacent cells on the board such that exactly one of the tilings has both squares occupied by a tile.)

Example:
Input: 3
Output: 5
Explanation: 
The five different ways are listed below, different letters indicates different tiles:
XYZ XXZ XYY XXY XYY
XYZ YYZ XZZ XYY XXY

Note:

• N  will be in range [1, 1000].

Title:

There are two kinds of floor tiles in L-shape and I-shape. How many ways are there to make 2*N floor tiles.

Ideas:

It can be said to be very pitiful. At first glance, I naively thought that dp[i]=dp[i-1]+dp[i-2]+2*dp[i-3], essentially ignoring a special case, namely LIL, where I is the horizontal of that situation. So use a two-dimensional dp array, dp[i]][0] means that the i-th column is full, and dp[i][1] means that the i-th column has a block. Then draw the graph to get the state transition equation, be careful not to double count.

Code:

class Solution {
    public int numTilings(int N) {
        if(N==1)
            return 1;
        if(N==2)
            return 2;
        if(N==3)
            return 5;
        long [][]dp=new long [N+1][2];
        for(int i=0;i<=N;i++)
        {
            dp[i][0]=0;
            dp[i][1]=0;
        }
        dp[1][0]=1;
        dp[1][1]=0;
        dp[2][0]=2;
        dp[2][1]=2;
        for(int i=3;i<=N;i++)
        {
            dp[i][0]=(int)((dp[i-1][0]+dp[i-1][1]+dp[i-2][0])%(1e9+7));
            dp[i][1]=(int)((dp[i-2][0]*2+dp[i-1][1])%(1e9+7));
        }
        return (int)dp[N][0];
    }
}

PS: This question also card data type. . . .