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.

Given n=3, k=2 return 6

      post 1,   post 2, post 3way1    0         0       1 way2    0         1       0way3    0         1       1way4    1         0       0way5    1         0       1way6    1         1       0

 1 class Solution { 2 public: 3     /** 4      * @param n non-negative integer, n posts 5      * @param k non-negative integer, k colors 6      * @return an integer, the total number of ways 7      */ 8     int numWays(int n, int k) { 9         // Write your code here10         11         // dp[i] indicates the number of ways of painting until i12         // If the color at i is different from its former fence, then there are (k - 1) choices times dp[i - 1]; If the color at i is the same from its former fence, then it must be different from (i - 2)th fence, so there are (k - 1)  * dp[i - 2] choices.13         vector<int> dp = {0, k, k * k};14         int index = 3;15         while (index <= n) 16             dp.push_back((k - 1) * (dp[index - 1] + dp[index++ - 2]));17         return dp[n];18     }19 };