There are a row of n houses, each house can be painted with one of the three colors: red, blue or green. 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 3 cost matrix. For example, costs[0][0] is the cost of painting house 0 with color red; costs[1][2] is the cost of painting house 1 with color green, and so on... Find the minimum cost to paint all houses.
Note:
All costs are positive integers.
此题和house robber有点像，也是用动态规划的做法做，一般做动态规划的时候，做一个与给定参数数组规模相同的数组作为dp数组，然后写出状态方程，再思考一下初始条件就可以了，代码如下：
public class Solution {
public int minCost(int[][] costs) {
if(costs==null||costs.length==0) return 0;
int m = costs.length;
int[][] dp = new int[m][3];
if(costs.length==1) return Math.min(costs[0][0],Math.min(costs[0][1],costs[0][2]));
dp[0][0] = costs[0][0];
dp[0][1] = costs[0][1];
dp[0][2] = costs[0][2];
for(int i=1;i<m;i++){
dp[i][0] = Math.min(dp[i-1][1],dp[i-1][2])+costs[i][0];
dp[i][1] = Math.min(dp[i-1][0],dp[i-1][2])+costs[i][1];
dp[i][2] = Math.min(dp[i-1][0],dp[i-1][1])+costs[i][2];
}
return Math.min(dp[m-1][0],Math.min(dp[m-1][1],dp[m-1][2]));
}
}