Weekly Contest 73 leetcode 790. Domino and Tromino Tiling

We have two types of tiles(地砖): 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.

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]

.

这道题很明显是一道 DP 题。因此关键在于找到 DP 的递归公式。
方法是找规律。

序列如下: 

1, 1, 2, 5, 11, 24, 53, 117, 258, 569, 1255

我们发现:

5 = 2 * 2 + 1

11 = 5 * 2 + 1

24 = 11 * 2 + 2

53 = 24 * 2 + 5

117 = 57 * 2 + 11

A
= A[N-1] * 2 + A[N-3]

使用DP的思路来考虑，为什么会出现这个公式呢？

A
can be get from A[N-1] by simply add

X

X

to the right of A[N-1],

A
can be get from A[N-2] by simply add
XX
YY
to the right of A[N-2] , Note that the method must be unique from those from A[N-1], thus the following tilling is not allowed:
XY
XY
cause it will be replicate to the cases from A[N-1] to A

How about A[N-3] and all previous cases? There are always 2 unique ways to turn them into A
(add the following blocks to the right of A[N-i]):
1.
XXY (from A[N-3])
XYY
XXYY (from A[N-4])
XZZY
XXZZY (from A[N-5])
XZZYY
XXZZYY (from A[N-6])
XZZZZY
etc…
2. vertically flip case 1(垂直反转case1)
Thus:
A
= A[N-1] + A[N-2] + 2 * sum(A[0:N-2])



重点在于：We can find dp[n-1] and dp[n-2] have one way to translate to dp
, and dp[n-3] ... dp[0] have two ways to translate to dp
. 然后就是推导公式，化简过程。

package leetcode;

public class Domino_and_Tromino_Tiling_790 {

public int numTilings(int N) {
int mod=1000000007;
long[] dp=new long[1001];
dp[1]=1;
dp[2]=2;
dp[3]=5;
for(int i=4;i<=N;++i){
dp[i]=2*dp[i-1]+dp[i-3];
dp[i]=dp[i]%mod;
}
return (int)dp
;
}

public static void main(String[] args) {
// TODO Auto-generated method stub
Domino_and_Tromino_Tiling_790 d=new Domino_and_Tromino_Tiling_790();
System.out.println(d.numTilings(30));
}
}