Leetcode 276: Paint Fence

最新推荐文章于 2019-02-21 10:14:12 发布

转载最新推荐文章于 2019-02-21 10:14:12 发布 · 88 阅读

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原文链接：http://www.cnblogs.com/liangmou/p/8054810.html

本文介绍了一种使用动态规划解决的计算机科学问题：如何给有n个柱子的围栏涂上k种颜色，使得任意相邻两个柱子的颜色不同或最多相同一次。文章提供了两种解决方案的Java代码实现。

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.

 1 public class Solution {
 2     public int NumWays(int n, int k) {
 3         if (n < 1 || k < 1) return 0;
 4         if (n == 1) return k;
 5         
 6         int dpSame = k, dpDifferent = k * (k - 1);
 7         
 8         for (int i = 2; i < n; i++)
 9         {
10             var tmp = dpSame;
11             dpSame = dpDifferent;
12             dpDifferent = (tmp + dpDifferent) * (k - 1); 
13         }
14         
15         return dpSame + dpDifferent;
16     }
17     
18     private int DFS(int n, int k, int start, int cur, bool sameColor)
19     {
20         if (start >= n)
21         {
22             return cur;
23         }
24         
25         if (sameColor)
26         {
27             return DFS(n, k, start + 1, cur * (k - 1), false);
28         }
29         else
30         {
31             return DFS(n, k, start + 1, cur, true) + DFS(n, k, start + 1, cur * (k - 1), false);
32         }
33     }
34 }
35 
36 // DFS, timeout
37 public class Solution1 {
38     public int NumWays(int n, int k) {
39         if (n < 1 || k < 1) return 0;
40         
41         return DFS(n, k, 1, k, false);
42     }
43     
44     private int DFS(int n, int k, int start, int cur, bool sameColor)
45     {
46         if (start >= n)
47         {
48             return cur;
49         }
50         
51         if (sameColor)
52         {
53             return DFS(n, k, start + 1, cur * (k - 1), false);
54         }
55         else
56         {
57             return DFS(n, k, start + 1, cur, true) + DFS(n, k, start + 1, cur * (k - 1), false);
58         }
59     }
60 }