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.
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.

  R.   b  G 
0 [[17,2,17]
1 [16,16,5]
2 [14,3,19]]



                
https://leetcode.com/problems/paint-house/discuss/68203/Share-my-very-simple-Java-solution-with-explanation

The basic idea is when we have painted the first i houses, and want to paint the i+1 th house, we have 3 choices: paint it either red, or green, or blue. If we choose to paint it red, we have the follow deduction:

paintCurrentRed = min(paintPreviousGreen,paintPreviousBlue) + costs[i+1][0]
Same for the green and blue situation. And the initialization is set to costs[0], so we get the code:

public class Solution {
public int minCost(int[][] costs) {
    if(costs.length==0) return 0;
    int lastR = costs[0][0];
    int lastG = costs[0][1];
    int lastB = costs[0][2];
    for(int i=1; i<costs.length; i++){
        int curR = Math.min(lastG,lastB)+costs[i][0];
        int curG = Math.min(lastR,lastB)+costs[i][1];
        int curB = Math.min(lastR,lastG)+costs[i][2];
        lastR = curR;
        lastG = curG;
        lastB = curB;
    }
    return Math.min(Math.min(lastR,lastG),lastB);
}