[LC] 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.

class Solution {
    public int minCost(int[][] costs) {
        if (costs == null || costs.length == 0 || costs[0].length == 0) {
            return 0;
        }
        int curRed = costs[0][0];
        int curBlue = costs[0][1];
        int curGreen = costs[0][2];
        int prevRed = curRed;
        int prevBlue = curBlue;
        int prevGreen = curGreen;
        for (int i = 1; i < costs.length; i++) {
            curRed = Math.min(prevBlue, prevGreen) + costs[i][0];
            curBlue = Math.min(prevRed, prevGreen) + costs[i][1];
            curGreen = Math.min(prevRed, prevBlue) + costs[i][2];
            prevRed = curRed;
            prevBlue = curBlue;
            prevGreen = curGreen;
        }
        return Math.min(curRed, Math.min(curBlue, curGreen));
    }
}

原文地址：https://www.cnblogs.com/xuanlu/p/12053778.html