# 276. Paint Fence 栅栏涂色

@TOC

## # 题目描述

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.

Example:

``````Input: n = 3, k = 2
Output: 6
Explanation: Take c1 as color 1, c2 as color 2. All possible ways are:

post1  post2  post3
-----      -----  -----  -----
1         c1     c1     c2
2         c1     c2     c1
3         c1     c2     c2
4         c2     c1     c1
5         c2     c1     c2
6         c2     c2     c1
``````

## # 解题方法

### # 动态规划

1. 和prev相同，此时说明prev的颜色应与pprev的颜色不同，pprev的涂色方法有 F(n - 2) 种，prev的涂色方式有 (k - 1) 种，所以此时情况应为 F(n - 2) * (k - 1)
2. 和prev不同，prev的涂色方法有 F(n - 1) 种，cur的涂色方式有 (k - 1) 种，此时情况应为 F(n - 1) * (k - 1)

C++代码如下：

``````class Solution {
public:
int numWays(int n, int k) {
if (n == 0) return 0;
if (n == 1) return k;
int pprev = k;
int prev = k * k;
int cur = k * k;
for (int i = 2; i < n; ++i) {
cur = pprev * (k - 1) + prev * (k - 1);
pprev = prev;
prev = cur;
}
return cur;
}
};
``````

## # 日期

2019 年 9 月 18 日 —— 今日又是九一八