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() == 0 || costs[0].size() == 0) return 0;
        int N = costs.size();
        int K = costs[0].size();
        vector<vector<int>> dp(N, vector<int>(K, 0));
        for (int i = 0; i < K; i++) dp[0][i] = costs[0][i];
        for (int i = 1; i < N; i++) {
            int minval1 = INT_MAX, minval2 = INT_MAX;
            int minidx1 = 0, minidx2 = 0;
            for (int j = 0; j < K; j++) {
                if (minval1 > dp[i-1][j]) {
                    minval1 = dp[i-1][j];
                    minidx1 = j;
                }
            }
            for (int j = 0; j < K; j++) {
                if (minval2 > dp[i-1][j] && j != minidx1) {
                    minval2 = dp[i-1][j];
                    minidx2 = j;
                }
            }
            for (int j = 0; 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 = 0; i<K; i++) {
            minval = min(minval, dp[N-1][i]);
        }
        return minval;
    }
};