题目大意：有两种不同形状的骨牌(1×2长条形，L型）无限多块。给你一个2xN的板子，问一共有多少不同的方式可以完全覆盖。

We have two types of tiles: a 2×1 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

Idea: DP

dp[i][0]: ways to cover i cols, both rows of i-th col are covered
dp[i][1]:  ways to cover i cols, only top row of i-th col is covered
dp[i][2]:  ways to cover i cols, only bottom row of i-th col is covered

Solution 1: DP

Time complexity: O(N)

Space complexity: O(N)

// Author: Huahua
// Running time: 4 ms
class Solution {
public:
  int numTilings(int N) {
    constexpr int kMod = 1000000007;
    vector<vector<long>> dp(N + 1, vector<long>(3, 0));    
    dp[0][0] = dp[1][0] = 1;
    for (int i = 2; i <= N; ++i) {
      dp[i][0] = (dp[i - 1][0] + dp[i - 2][0] + dp[i - 1][1] + dp[i - 1][2]) % kMod;
      dp[i][1] = (dp[i - 2][0] + dp[i - 1][2]) % kMod;
      dp[i][2] = (dp[i - 2][0] + dp[i - 1][1]) % kMod;
    }
    
    return dp[N][0];
  }
};

Since dp[i][1] always equals to dp[i][2], we can simplify a bit.

// Author: Huahua
// Running time: 4 ms
class Solution {
public:
  int numTilings(int N) {
    constexpr int kMod = 1000000007;
    vector<vector<long>> dp(N + 1, vector<long>(2, 0));    
    dp[0][0] = dp[1][0] = 1;
    for (int i = 2; i <= N; ++i) {
      dp[i][0] = (dp[i - 1][0] + dp[i - 2][0] + 2 * dp[i - 1][1]) % kMod;
      dp[i][1] = (dp[i - 2][0] + dp[i - 1][1]) % kMod;      
    }
    
    return dp[N][0];
  }
};

Solution 2: DP

Another way to think about this problem

define: dp[i] ways to completely covert the i*2 board.

dp[0] = 1 # {}
dp[1] = 1 # {|}
dp[2] = 2 # {||, =}
dp[3] = 5 # {|||, |=, =|, ⌊⌉, ⌈⌋} = dp[2] ⊗ {|} + dp[1] ⊗ {=} + dp[0] ⊗ {⌊⌉, ⌈⌋}
dp[4] = 11 # dp[3] ⊗ {|} + dp[2] ⊗ {=} + dp[1] ⊗ {⌊⌉, ⌈⌋} + dp[0] ⊗ {⌊¯⌋,⌈_⌉}
dp[5] = 24 # dp[4] ⊗ {|} + dp[3] ⊗ {=} + 2*(dp[2] + dp[1] + dp[0])
...
dp[n] = dp[n-1] + dp[n-2] + 2*(dp[n-3] + ... + dp[0])
      = dp[n-1] + dp[n-3] + [dp[n-2] + dp[n-3] + 2*(dp[n-4] + ... + dp[0])]
      = dp[n-1] + dp[n-3] + dp[n-1]
      = 2*dp[n-1] + dp[n-3]

// Author: Huahua
// Running time: 3 ms
class Solution {
public:
  int numTilings(int N) {
    constexpr int kMod = 1000000007;
    vector<long> dp(N + 1, 1);
    dp[2] = 2;
    for (int i = 3; i <= N; ++i)
      dp[i] = (dp[i - 3] + dp[i - 1] * 2) % kMod;
    return dp[N];
  }
};

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