256. Paint House - Easy

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.

 

dynamic programming

在costs的基础上，每一行每个元素累加与上一次不同颜色的cost，在最后一行取最小值

time: O(m), space: O(1)　　-- m:  # rows of costs

class Solution {
    public int minCost(int[][] costs) {
        if(costs == null || costs.length == 0 || costs[0].length == 0) return 0;
        
        for(int i = 1; i < costs.length; 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]);
        }

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