[Leetcode] 790. Domino and Tromino Tiling 解题报告_dominoandtrominotiling-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/79744561

这篇博客详细介绍了LeetCode第790题的解题思路，即如何使用2x1的多米诺骨牌和L形 tromino 形状来铺满2xN的棋盘，并讨论了动态规划的解决方案。博主给出了初始状态和递推关系，并逐步优化了动态规划的空间和时间复杂度，最终简化了代码实现。

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题目：

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].

思路：

我们定义fe[n]表示2*n的格子有多少种铺法，定义fo[n]表示2*n + 1的格子有多少种铺法（也就是在2*n的格子上面含有一个格子），然后推导fe和fo的递推关系。对于fe的最顶层，我们分别有如下图所示的几种铺法，因此，可以得出递推关系为：

fe[n] = fe[n - 1] + fe[n - 2] + 2fe[n - 3] + 2fo[n - 3];

fo[n] = fo[n - 1] + fe[n - 1].

特别地，我们容易得知：fe[1] = 1, fe[2] = 2, fe[3] = 5, fo[1] = 2, fo[2] = 2, fo[3] = 4。因此就可以根据初始条件和递推关系写出基于动态规划的源代码了。

我们在下面写出的源代码的空间复杂度为O(N)，时间复杂度为O(N)。但是注意到fe[n]和fo[n]也仅仅只和fe[n-1], fe[n-2], fe[n-3]以及fo[n-1], fo[n-2], fo[n-3]有关，所以还可以进一步将空间复杂度从O(n)优化到O(1)。读者可以自行实现了^_^。

更新：后来发现上面的递推公式还是推导复杂了，其实把fo(n) = fo(n - 1) + fe(n - 1)代入fe(n),可以得到更简单的递推公式：fe[n] = fe[n - 1] + fe[n - 2] + 2fo[n-2]，甚至还可以进一步优化为fe[n] = fe[n - 1] + fo[n - 1] + fo[n-2]，还可以进一步优化为fe[n] = fo[n] + fo[n - 2]。这样写出来的代码应该就更简洁优雅了。

代码：

原有代码：

class Solution {
public:
    int numTilings(int N) {
        vector<long long> fe(max(4, N + 1), 0), fo(max(4, N + 1), 0);
        long long mod = 1000000007;
        fe[1] = 1, fe[2] = 2, fe[3] = 5;
        fo[1] = 1, fo[2] = 2, fo[3] = 4;
        for (int i = 4; i <= N; ++i) {
            fe[i] = fe[i - 1] + fe[i - 2] + 2 * fe[i - 3] + 2 * fo[i - 3];
            fo[i] = fo[i - 1] + fe[i - 1];
            fe[i] %= mod;
            fo[i] %= mod;
        }
        return fe[N];
    }
};

更新后的代码：

class Solution {
public:
    int numTilings(int N) {
        vector<long long> fe(max(3, N + 1), 0), fo(max(3, N + 1), 0);
        long long mod = 1000000007;
        fe[1] = 1, fe[2] = 2;
        fo[1] = 1, fo[2] = 2;
        for (int i = 3; i <= N; ++i) {
            fo[i] = fo[i - 1] + fe[i - 1];
            fe[i] = fo[i] + fo[i - 2];
            fe[i] %= mod;
            fo[i] %= mod;
        }
        return fe[N];
    }
};