LeetCode 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;
     }
 }