LWC 73: 790. Domino and Tromino Tiling

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本文探讨了使用2x1多米诺骨牌和“L”型三米诺骨牌铺满2xN棋盘的方法数量，并提供了解决方案的动态规划思路及Java和Python代码实现。

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LWC 73: 790. Domino and Tromino Tiling

传送门：790. Domino and Tromino Tiling

Problem:

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].

思路：
动态规划，切分子问题。比如考虑N等于1可能出现的状态：

这里写图片描述

从上至下对应状态为0， 1， 2， 3，因此可以得到N=2时，每个状态的转移方程：

dp[i][0] = dp[i - 1][0] + dp[i - 1][3] + dp[i - 2][1] + dp[i - 2][2]
dp[i][1] = dp[i - 1][0] + dp[i - 1][2]
dp[i][2] = dp[i - 1][0] + dp[i - 1][1]
dp[i][3] = dp[i - 1][0]

代码如下：

    public int numTilings(int N) {
        long[][] dp = new long[N+1][4];
        int mod = 1000000007;
        dp[1][0] = 1;
        dp[1][1] = 1;
        dp[1][2] = 1;
        dp[1][3] = 1;
        for (int i = 2; i <= N; ++i) {
            dp[i][0] = (dp[i - 1][0] + dp[i - 1][3] + dp[i - 2][1] + dp[i - 2][2]) % mod;
            dp[i][1] = (dp[i - 1][0] + dp[i - 1][2]) % mod;
            dp[i][2] = (dp[i - 1][0] + dp[i - 1][1]) % mod;
            dp[i][3] = dp[i - 1][0] % mod;
        }
        return (int)dp[N][0];
    }

Python版本：

class Solution(object):
    def numTilings(self, N):
        """
        :type N: int
        :rtype: int
        """
        dp = [[0] * 4 for _ in range(N + 1)]
        mod = 1000000007
        dp[1][0] = dp[1][1] = dp[1][2] = dp[1][3] = 1
        for i in range(2, N + 1):
            dp[i][0] = (dp[i - 1][0] + dp[i - 1][3] + dp[i - 2][1] + dp[i - 2][2]) % mod
            dp[i][1] = (dp[i - 1][0] + dp[i - 1][2]) % mod
            dp[i][2] = (dp[i - 1][0] + dp[i - 1][1]) % mod
            dp[i][3] = dp[i - 1][0] % mod
        return dp[N][0]