265 Paint House 2

There are a row ofnhouses, each house can be painted with one of thekcolors. 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 anxkcost 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.

Note:
All costs are positive integers.

Example:

Input:
 [[1,5,3],[2,9,4]]

Output:
 5

Explanation: 
Paint house 0 into color 0, paint house 1 into color 2. Minimum cost: 1 + 4 = 5; 
             Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5.

Follow up:
Could you solve it inO(nk) runtime?

class Solution {
    public int minCostII(int[][] costs) {
        if (costs == null || costs.length == 0) {
            return 0;
        }
        int n = costs.length;
        int k = costs[0].length;
        //f[i][j]
        int[][] f = new int[n + 1][k];
        int min1, min2;
        int idx1 = 0;
        int idx2 = 0;
        for (int i = 1; i <= n; i++) {
            min1 = Integer.MAX_VALUE;
            min2 = Integer.MAX_VALUE;;
            for (int j = 0; j < k; j++) {
                if(f[i - 1][j] < min1) {
                    min2 = min1;
                    idx2 = idx1;
                    min1 = f[i - 1][j];
                    idx1 = j;
                }else if(f[i - 1][j] < min2) {
                    min2 = f[i - 1][j];
                    idx2 = j;
                }
            }
            System.out.println("min1: "  + min1 + " min2: " + min2);
            for(int j = 0; j < k; j++) {
                if (j == idx1) {
                    if (min2 ==  Integer.MAX_VALUE) {
                         f[i][j] = costs[i - 1][j];
                    }else{
                        f[i][j] = min2 + costs[i - 1][j];
                    }

                }else {
                     if (min1 ==  Integer.MAX_VALUE) {
                         f[i][j] = costs[i - 1][j];
                    }else{
                        f[i][j] = min1 + costs[i - 1][j];
                    }

                }
            }
        }
        int res = Integer.MAX_VALUE;
        for (int i = 0; i < k; i++) {
            res = Math.min(res, f[n][i]);
        }
        return res;
    }
}