## 题目描述：

LeetCode 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]`.

## 解题思路：

dp[x][y]表示长度（两行的最小值）为x，末尾形状为y的拼接方法个数

y有三种可能：

```0表示末尾没有多余部分

1表示第一行多出1个单元格

2表示第二行多出1个单元格```

```dp[x][0] = (dp[x - 1][0] + sum(dp[x - 2])) % MOD   1个竖条， 2个横条，L7， rotate(L7)

dp[x][1] = (dp[x - 1][0] + dp[x - 1][2]) % MOD    rotate(L)，L + 第一行横条

dp[x][2] = (dp[x - 1][0] + dp[x - 1][1]) % MOD    L，rotate(L) + 第二行横条```

## Python代码：

``````class Solution(object):
def numTilings(self, N):
"""
:type N: int
:rtype: int
"""
MOD = 10**9 + 7
dp = [[0] * 3 for x in range(N + 10)]
dp[0] = [1, 0, 0]
dp[1] = [1, 1, 1]
for x in range(2, N + 1):
dp[x][0] = (dp[x - 1][0] + sum(dp[x - 2])) % MOD
dp[x][1] = (dp[x - 1][0] + dp[x - 1][2]) % MOD
dp[x][2] = (dp[x - 1][0] + dp[x - 1][1]) % MOD
return dp[N][0]
``````

Pingbacks已关闭。

1. DennyZhang 发布于 2018年2月27日 06:57 #

比较新颖的DP题。出题者比较赞。

经常看博主的解答，还有收获。能加个微信吗？

Email: contact at dennyzhang.com