790. Domino and Tromino Tiling
You have two types of tiles: a 2 x 1 domino shape and a tromino shape. You may rotate these shapes.
Given an integer n, return the number of ways to tile an 2 x n board. Since the answer may be very large, return it modulo 1 0 9 + 7 10^9 + 7 109+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 1:
Input: n = 3
Output: 5
Explanation: The five different ways are show above.
Example 2:
Input: n = 1
Output: 1
Constraints:
- 1 <= n <= 1000
From: LeetCode
Link: 790. Domino and Tromino Tiling
Solution:
Ideas：
- Define a recurrence relation to calculate the number of tilings for a board of width n.
- The base cases will be small widths for which we can manually count the number of tilings.
- For larger widths, we build up the solution from the base cases, using the recurrence relation.
- We need to consider the last column which could be filled by:
- A vertical domino, which leaves the subproblem of tiling a 2 x (n-1) board.
- Two horizontal dominos, which leaves the subproblem of tiling a 2 x (n-2) board.
- A tromino along with a domino, which will lead to two subproblems: tiling a 2 x (n-2) board and a 2 x (n-3) board.
- Since the answer can be very large, we will return it modulo 1 0 9 + 7 10^9+7 109+7.
Caode:
c
int numTilings(int n) {
if (n == 1) return 1;
if (n == 2) return 2;
if (n == 3) return 5;
long dp[n+1];
dp[0] = 1; dp[1] = 1; dp[2] = 2; dp[3] = 5;
for (int i = 4; i <= n; ++i) {
dp[i] = (2 * dp[i-1] % 1000000007 + dp[i-3]) % 1000000007; // Main recurrence relation
}
return (int) dp[n];
}