Description

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 nx 3cost matrix. For example,costs[0][0]is the cost of painting house 0with color red;costs[1][2]is the cost of painting house 1with color green, and so on... Find the minimum cost to paint all houses.

Example

Givencosts=[[14,2,11],[11,14,5],[14,3,10]]return10

house 0 is blue, house 1 is green, house 2 is blue,2 + 5 + 3 = 10

Solution

第i个房子上涂第j种颜色的结果取决于第i-1个房子上涂除了第j种颜色的最小值，即dp[i][j] = dp[i-1][k(k!=j)]
因为只有三种颜色，所以除了j的其他颜色直接枚举即可
直接用cost来存dp的值，节省额外内存空间

public class Solution {
    /**
     * @param costs n x 3 cost matrix
     * @return an integer, the minimum cost to paint all houses
     */
    public int minCost(int[][] costs) {
        // Write your code here
        if (costs == null || costs.length == 0 || costs[0] == null || costs[0].length == 0) {
            return 0;
        }

        int n = costs.length;
        for (int i = 1; i < n; i++) {
            costs[i][0] = costs[i][0] + Math.min(costs[i-1][1], costs[i-1][2]);
            costs[i][1] = costs[i][1] + Math.min(costs[i-1][0], costs[i-1][2]);
            costs[i][2] = costs[i][2] + Math.min(costs[i-1][0], costs[i-1][1]);
        }


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