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.

Note:
All costs are positive integers.

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

思路跟第一题类似，就是找上一层最小或者第二最小的数，如果和自己颜色相同，取另外一个。然后每层跟新原来的数组，跟上次一样。

public class Solution {    public int minCostII(int[][] costs) {        if(costs == null || costs.length == 0) return 0;        int preMin = 0; int preSec = 0; int preIndex = -1;        for(int i=0; i<costs.length; i++){            int curMin = Integer.MAX_VALUE;            int curSec = Integer.MAX_VALUE;            int curIndex = -1;            for(int j=0; j<costs[i].length; j++){                costs[i][j] = costs[i][j] + (j == preIndex ? preSec : preMin);                if(costs[i][j] < curMin) {                    curSec = curMin;                    curMin = costs[i][j];                    curIndex = j;                } else if(costs[i][j] < curSec){                    curSec = costs[i][j];                }            }            preMin = curMin;            preSec = curSec;            preIndex = curIndex;        }        return preMin;    }}