[Swift]LeetCode790. 多米诺和托米诺平铺 | Domino and Tromino Tiling

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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].

---

有两种形状的瓷砖：一种是 2x1 的多米诺形，另一种是形如 "L" 的托米诺形。两种形状都可以旋转。

XX  <- 多米诺

XX  <- "L" 托米诺
X

给定 N 的值，有多少种方法可以平铺 2 x N 的面板？返回值 mod 10^9 + 7。

（平铺指的是每个正方形都必须有瓷砖覆盖。两个平铺不同，当且仅当面板上有四个方向上的相邻单元中的两个，使得恰好有一个平铺有一个瓷砖占据两个正方形。）

示例:
输入: 3
输出: 5
解释: 
下面列出了五种不同的方法，不同字母代表不同瓷砖：
XYZ XXZ XYY XXY XYY
XYZ YYZ XZZ XYY XXY

提示：

• N  的范围是 [1, 1000]

---

Runtime: 4 ms

Memory Usage: 19 MB

class Solution {
    func numTilings(_ N: Int) -> Int {
        switch N
        {
            case 0:
            return 1
            case 1,2:
            return N
            default:
            break
        }        
        var M:Int = Int(1e9) + 7
        var dp:[Int] = [Int](repeating:0,count:N + 1)
        dp[0] = 1
        dp[1] = 1
        dp[2] = 2
        for i in 3...N
        {
            dp[i] = (dp[i - 1] * 2 + dp[i - 3]) % M
        }      
        return dp[N]
    }
}

---

4ms

class Solution {
    func numTilings(_ N: Int) -> Int {
        let MOD = 1000000007
        if N == 1 { return 1 }
        if N == 2 { return 2 }
        var dp = [Int](repeating: 0, count: N+1)
        dp[0] = 1; dp[1] = 1; dp[2] = 2
        for i in 3...N {
            dp[i] = (2*dp[i-1] + dp[i-3])%MOD
        }
        return dp[N]
    }
}