LC 265. Paint House II

There are a row of n houses, each house can be painted with one of the k colors. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color.

The cost of painting each house with a certain color is represented by a n x k cost matrix. For example, costs[0][0] is the cost of painting house 0 with color 0; costs[1][2] is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.

思路：DP，第n个house如果paint color i, 那第n-1个house就不能。

注意minval1和minval2是前一次dp的最小值，而不是前一个costs的最小值。

 class Solution {
 public:
     int minCostII(vector<vector<int>>& costs) {
         if (costs.size() ==  || costs[].size() == ) return ;
         int N = costs.size();
         int K = costs[].size();
         vector<vector<int>> dp(N, vector<int>(K, ));
         for (int i = ; i < K; i++) dp[][i] = costs[][i];
         for (int i = ; i < N; i++) {
             int minval1 = INT_MAX, minval2 = INT_MAX;
             int minidx1 = , minidx2 = ;
             for (int j = ; j < K; j++) {
                 if (minval1 > dp[i-][j]) {
                     minval1 = dp[i-][j];
                     minidx1 = j;
                 }
             }
             for (int j = ; j < K; j++) {
                 if (minval2 > dp[i-][j] && j != minidx1) {
                     minval2 = dp[i-][j];
                     minidx2 = j;
                 }
             }
             for (int j = ; j < K; j++) {
                 if (minidx1 == j) dp[i][j] = costs[i][j] + minval2;
                 else dp[i][j] = costs[i][j] + minval1;
             }
         }
         int minval = INT_MAX;
         for (int i = ; i<K; i++) {
             minval = min(minval, dp[N-][i]);
         }
         return minval;
     }
 };