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.

这道题是之前那道Paint House的拓展，那道题只让用红绿蓝三种颜色来粉刷房子，而这道题让我们用k种颜色，这道题不能用之前那题的解法，会TLE。这题的解法的思路还是用DP，但是在找不同颜色的最小值不是遍历所有不同颜色，而是用min1和min2来记录之前房子的最小和第二小的花费的颜色，如果当前房子颜色和min1相同，那么我们用min2对应的值计算，反之我们用min1对应的值，这种解法实际上也包含了求次小值的方法，感觉也是一种很棒的解题思路，参见代码如下：

public class Solution {    public int minCostII(int[][] costs) {        if(costs == null || costs.length == 0 || costs[0]. length == 0){              return 0;          }          int min1 = 0, min2 = 0, preIndex = -1;        for(int i=0; i<costs.length; i++){            int m1 = Integer.MAX_VALUE, m2 = Integer.MAX_VALUE, curIndex = -1; //curIndex一定要保留，不然会在下面被冲掉            for(int j=0; j<costs[0].length; j++){                int cost = costs[i][j] + (j == preIndex ? min2: min1); //不保留的话这里就被冲掉了                if(m1 > cost){ //求第二小的元素的方法                    m2 = m1;                    m1 = cost;                    curIndex = j;                }else if(m2 > cost){                    m2 = cost;                }            }            min1 = m1;             min2 = m2;            preIndex = curIndex;        }        return min1;    }}