[LeetCode] Paint Fence

最新推荐文章于 2025-08-08 22:01:10 发布

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最新推荐文章于 2025-08-08 22:01:10 发布

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文章标签：数据结构与算法

本文介绍了一种动态规划算法，用于解决给定长度和颜色数量的围栏涂色问题，确保相邻围栏颜色不同，详细解释了算法原理及其实现过程。

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Paint Fence

There is a fence with n posts, each post can be painted with one of the k colors.

You have to paint all the posts such that no more than two adjacent fence posts have the same color.

Return the total number of ways you can paint the fence.

Note:
n and k are non-negative integers.

动态规划，可以使用滚动数组节省空间。dp[i][j][0]表示第i个fence的颜色为j，同时与第i-1个fence颜色不同的方案数；dp[i][j][1]表示第i个fence的颜色为j，同时与第i-1个fence颜色相同的方案数。

 1 class Solution {
 2 public:
 3     int numWays(int n, int k) {
 4         if (n == 0 || k == 0) return 0;
 5         if (n == 1) return k;
 6         int dp[2][k][2];
 7         for (int j = 0; j < k; ++j) {
 8             dp[0][j][0] = 1;
 9             dp[0][j][1] = 0;
10         }
11         for (int i = 1; i < n; ++i) {
12             int sum = 0;
13             for (int j = 0; j < k; ++j) {
14                 sum += (dp[(i-1)%2][j][0] + dp[(i-1)%2][j][1]);
15             }
16             for (int j = 0; j < k; ++j) {
17                 dp[i%2][j][1] = dp[(i-1)%2][j][0];
18                 dp[i%2][j][0] = sum - dp[(i-1)%2][j][0] - dp[(i-1)%2][j][1];
19             }
20         }
21         int res = 0;
22         for (int i = 0; i < k; ++i) {
23             res += dp[(n-1)%2][i][0] + dp[(n-1)%2][i][1];
24         }
25         return res;
26     }
27 };