256. Paint House

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.

这道题用动态规划解题。三种颜色代表三种状态，red, blue,green。第i座房子涂red, blue,green的最小消耗是其他两种颜色i-1的值加上当前房子的cost。代码如下：

public class Solution {    public int minCost(int[][] costs) {        int n = costs.length;        if (n == 0) {            return 0;        }        int[] red = new int[n + 1];        int[] blue = new int[n + 1];        int[] green = new int[n + 1];        for (int i = 1; i <= n; i ++) {            red[i] = Math.min(blue[i - 1] + costs[i - 1][0], green[i - 1] + costs[i - 1][0]);            blue[i] = Math.min(red[i - 1] + costs[i - 1][1], green[i - 1] + costs[i - 1][1]);            green[i] = Math.min(red[i - 1] + costs[i - 1][2], blue[i - 1] + costs[i - 1][2]);        }        return Math.min(red[n], Math.min(blue[n], green[n]));    }}