790. Domino and Tromino Tiling-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_36718317/article/details/79791825

探讨使用L字形和I字形两种地板砖拼接2xN大小的矩形区域的方法数量，并通过动态规划解决该问题。文章提供了一个Java实现方案，详细解释了状态转移方程。

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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].

题意：

有L字形和I字形的两种地板砖，问拼成2*N的地板砖有多少种方法。

思路：

可以说是很坑的了。咋一看天真地以为dp[i]=dp[i-1]+dp[i-2]+2*dp[i-3]，实质上忽略了一种特殊情况，即LIL,其中I是横着的那种情况。所以用一个二维dp数组，dp[i]][0]表示第i列铺满，dp[i][1]表示第i列有一块。然后画图可得状态转移方程，注意不要重复计数。

代码：

class Solution {
    public int numTilings(int N) {
        if(N==1)
            return 1;
        if(N==2)
            return 2;
        if(N==3)
            return 5;
        long [][]dp=new long [N+1][2];
        for(int i=0;i<=N;i++)
        {
            dp[i][0]=0;
            dp[i][1]=0;
        }
        dp[1][0]=1;
        dp[1][1]=0;
        dp[2][0]=2;
        dp[2][1]=2;
        for(int i=3;i<=N;i++)
        {
            dp[i][0]=(int)((dp[i-1][0]+dp[i-1][1]+dp[i-2][0])%(1e9+7));
            dp[i][1]=(int)((dp[i-2][0]*2+dp[i-1][1])%(1e9+7));
        }
        return (int)dp[N][0];
    }
}

PS：本题还卡数据类型。。。。