LWC 73: 790. Domino and Tromino Tiling
LWC 73: 790. Domino and Tromino Tiling
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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: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]