LWC 73: 790. Domino and Tromino Tiling

用户1147447

发布于 2019-05-26 09:41:31

5550

发布于 2019-05-26 09:41:31

文章被收录于专栏：机器学习入门

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:undefined 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：

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]

本文参与腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。

原始发表：2018年02月26日，如有侵权请联系 cloudcommunity@tencent.com 删除

python