### 题目

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


### 递归

1. 和前一条颜色不同: k-1种可能
2. 和前一条颜色相同: 1中可能

f(n) = f(n-1) * (k-1) + f(n-2) * (k-1)

#### 代码

class Solution {
public int numWays(int n, int k) {
if (n == 0 || k == 0) { return 0; }
if (n == 1) { return k; }
if (n == 2) { return k * k; }
return (numWays(n-1,k) + numWays(n-2,k)) * (k-1);
}
}


### 带表格的动态规划

#### 代码

public int numWays(int n, int k) {
int[] memo = new int[n+1];
return helper(n,k,memo);
}
private int helper(int n, int k, int[] memo) {
if (n == 0 || k == 0) { return 0; }
if (n == 1) { return k; }
if (n == 2) { return k * k; }
memo[n] = (((memo[n-1] == 0)? helper(n-1,k,memo) : memo[n-1]) + ((memo[n-2] == 0)? helper(n-2,k,memo) : memo[n-2])) * (k-1);
return memo[n];
}


### 自底向上的动态规划

#### 代码

public int numWays(int n, int k) {
if (n == 0 || k == 0) { return 0; }
if (n == 1) { return k; }
if (n == 2) { return k * k; }
int a = k, b = k * k;
int m = 3;
while (m++ <= n) {
int next = (a + b)*(k-1);
a = b;
b = next;
}
return b;
}