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.

Example:

Input: [[17,2,17],[16,16,5],[14,3,19]]

Output: 10

Explanation: Paint house 0 into blue, paint house 1 into green, paint house 2 into blue.

             Minimum cost: 2 + 5 + 3 = 10.

题意：

一排有n套房子，每个房子可以选一种颜色（红色、蓝色或者绿色）来粉刷。由于每个房子涂某种颜色的花费不同，求相邻房子不同色的粉刷最小花费。

Solution1： DP

scan一遍n套房子：

costs[i][0]               =      costs[i][0]                   +             min(costs[i-1][1],  costs[i-1][2])

当下选0号色的花费(和)     =  当下选0号色的花费(单价)    +      前套房子选1号色或者选2号色中较小花费

code

 class Solution {

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

         if(costs == null || costs.length == 0 || costs[0].length < 3) return 0;

         for(int i = 1; i < costs.length; i++ ){ // 照顾[i-1]所以 i 从 1 开始， 则 i - 1 从 0 开始。若从0开始，i 就只能取到costs.length-1

             costs[i][0] += Math.min(costs[i-1][1], costs[i-1][2]);

             costs[i][1] += Math.min(costs[i-1][0], costs[i-1][2]);

             costs[i][2] += Math.min(costs[i-1][1], costs[i-1][0]);

         }

         int n = costs.length-1;

         return Math.min(Math.min(costs[n][0], costs[n][1]), costs[n][2]);

     }

 }