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.

Note:
All costs are positive integers.

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

 

此题是paint house的升级，和paint house不同之处在于，每个房子粉刷的颜色不再是常数了，而是k，此题想了很久没有什么思路，便直接看大神的解题思路了。

首先定义两个最小值，一个是第一小的值，一个是第二小的值，代表遍历到每个房子各个颜色的成本最小值，和第二小的值。在粉刷下一个房子的时候，如果之前的房子所包含的颜色不在最小成本值里面，就把最小成本值加上去，如果在最小成本值里面，则把第二小的值加进去。代码如下：

public class Solution {

    public int minCostII(int[][] costs) {

        if(costs==null||costs.length==0||costs[0].length==0) return 0;

        int min1 = -1;

        int min2 = -1;

        int m = costs.length;

        int k = costs[0].length;

        for(int i=0;i<m;i++){

            int last1= min1;

            int last2 = min2;

            min1 = -1;

            min2 = -1;

            for(int j=0;j<k;j++){

                if(last1!=j){

                    costs[i][j]+=last1<0?0:costs[i-1][last1];

                }else{

                    costs[i][j]+=last1<0?0:costs[i-1][last2];

                }

                if(min1<0||costs[i][j]<costs[i][min1]){

                    min2 = min1;

                    min1 = j;

                }else if(min2<0||costs[i][j]<costs[i][min2]){

                    min2 = j;

                }

            }

        }

        return costs[m-1][min1];

    }

}