[leetcode]265. Paint House II粉刷房子(K色可选)

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.

Example:

Input: [[1,5,3],[2,9,4]]
Output: 5
Explanation: Paint house 0 into color 0, paint house 1 into color 2. Minimum cost: 1 + 4 = 5;
             Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5. 

题意：

一排有n套房子，每个房子可以选一种颜色（K种不同颜色）来粉刷。由于每个房子涂某种颜色的花费不同，求相邻房子不同色的粉刷最小花费。

做这个题的时候，想起跟老妈一起去过的Santa Cruz海棠花节的时候去过的这个网红海滩。

所以这个题，对于景区想打造网红般的colorful houses来讲，还是蛮有显示意义。

Solution1: DP

code

 class Solution {
     public int minCostII(int[][] costs) {
         // sanity check
         if (costs == null || costs.length == 0) return 0;

         //init NO.1 house's curMin1, curMin2
         int curMin1 = -1; // most minimum
         int curMin2 = -1; // second minimum
         for (int j = 0; j < costs[0].length; j++) {
             if (curMin1 < 0 || costs[0][j] < costs[0][curMin1]) {
                 curMin2 = curMin1;
                 curMin1 = j;
             } else if (curMin2 < 0 || costs[0][j] < costs[0][curMin2]) {
                 curMin2 = j;
             }
         }
         // scan from NO.2 house
         for (int i = 1; i < costs.length; i++) {  // scan n houses
             // get NO.1 house's curMin1, curMin2
             int preMin1 = curMin1;
             int preMin2 = curMin2;
             // init NO.2 house's curMin1, curMin2
             curMin1 = -1;
             curMin2 = -1;
             // try k colors
             for (int j = 0; j < costs[0].length; j++) {
                 if (j != preMin1) {
                     costs[i][j] += costs[i - 1][preMin1];
                 } else {
                     costs[i][j] += costs[i - 1][preMin2];
                 }
                 // update NO.2 house's curMin1, curMin2
                 if (curMin1 < 0 || costs[i][j] < costs[i][curMin1]) {
                     curMin2 = curMin1;
                     curMin1 = j;
                 } else if (curMin2 < 0 || costs[i][j] < costs[i][curMin2]) {
                     curMin2 = j;
                 }
             }
         }
         return costs[costs.length - 1][curMin1];
     }
 }