# 790     Domino and Tromino Tiling

## 790. [Domino and Tromino Tiling](https://leetcode.com/problems/domino-and-tromino-tiling/description/)

## 1. Question

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]`.

## 2. Implementation

**(1) DP**

思路: 从n = 1到n = 5先画出能拼出tile的，然后推导出递归公式。具体推导可以看[这里](http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-790-domino-and-tromino-tiling/)。数学推导可以参考[这个](https://leetcode.com/problems/domino-and-tromino-tiling/discuss/116581/Detail-and-explanation-of-O%28n%29-solution-why-dpn2*dn-1+dpn-3)。

```java
class Solution {
    public int numTilings(int N) {
        int MOD = 1000000007;

        long[][] dp = new long[N + 1][3];

        dp[0][0] = 1;
        dp
        [1][0] = 1;

        for (int i = 2; i <= N; i++) {
            dp[i][0] = (dp[i - 1][0] + dp[i - 2][0] + dp[i - 1][1] + dp[i - 1][2]) % MOD;
            dp[i][1] = (dp[i - 2][0] + dp[i - 1][2]) % MOD;
            dp[i][2] = (dp[i - 2][0] + dp[i - 1][1]) % MOD;
        }
        return (int)dp[N][0];
    }
}
```

## 3. Time & Space Complexity

**DP:**&#x65F6;间复杂度O(n), 空间复杂度O(n)


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