<p>You have two types of tiles: a <code>2 x 1</code> domino shape and a tromino shape. You may rotate these shapes.</p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/15/lc-domino.jpg" style="width: 362px; height: 195px;" />
<p>Given an integer n, return <em>the number of ways to tile an</em> <code>2 x n</code> <em>board</em>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>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.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/15/lc-domino1.jpg" style="width: 500px; height: 226px;" />
<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 5
<strong>Explanation:</strong> The five different ways are show above.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>

class Solution {
public:
    vector<int> memo;
    const int MOD = 1e9 + 7;  // 模数，防止溢出

    int numTilings(int n) 
    {
        memo = vector<int> (n + 1, -1);
        return dp(n);    
    }

    int dp(int n)
    {
        if (n == 1)
            return 1;
        if (n == 2)
            return 2;
        if (n == 3)
            return 5;  

        if (memo[n] != -1)
            return memo[n];

        int ans = (dp(n - 1) * 2 % MOD + dp(n - 3)) % MOD;
        memo[n] = ans;
        return ans;
    }
};

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