LeetCode 265. Paint House II

原题链接在这里：https://leetcode.com/problems/paint-house-ii/

题目：

There are a row of n houses, each house can be painted with one of the k colors. 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 k cost matrix. For example, costs[0][0] is the cost of painting house 0 with color 0; costs[1][2] is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.

Note:
All costs are positive integers.

题解：

类似Paint House. k中不同的颜色. paint当前house时, 若是当前的颜色k与之前最小cost的颜色不同时，加上之前的最小值. 若是当前k与之前相同就选之前给出第二小cost的颜色.

Time Complexity: O(kn). Space: O(1).

AC Java:

public class Solution {
    public int minCostII(int[][] costs) {
        if(costs == null || costs.length == 0){
            return 0;
        }
        int min1 = 0; //当前这个house涂完最小的cost
        int min2 = 0; //当前这个house涂完第二小的cost
        int lastColor = -1; //上个house选择的颜色
        for(int i = 0; i<costs.length; i++){
            int curMin1 = Integer.MAX_VALUE; //选择到当前颜色时的最小cost
            int curMin2 = Integer.MAX_VALUE; //选择到当前颜色时的第二小cost
            int curColor = -1;  
            for(int k = 0; k<costs[i].length; k++){
                int newCost = costs[i][k] + (k == lastColor ? min2 : min1);
                if(newCost < curMin1){
                    curMin2 = curMin1;
                    curMin1 = newCost;
                    curColor = k;
                }else if(newCost < curMin2){
                    curMin2 = newCost;
                }
            }
            min1 = curMin1;
            min2 = curMin2;
            lastColor = curColor;
        }
        return min1;
    }
}