Paint House

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.

Example

Given n = 3, k = 3, costs = [[14,2,11],[11,14,5],[14,3,10]] return10

house 0 is color 2, house 1 is color 3, house 2 is color 2, 2 + 5 + 3 = 10

分析：

这题和在一个矩阵上冲一个点走到另一个点所需的cost是一样的。

从第二个房子开始，找出使用和上一家不同颜色能够获取的最小的cost。最后一排就可以算出总的价格。

 1 public class Solution {
 2     /**
 3      * @param costs n x k cost matrix
 4      * @return an integer, the minimum cost to paint all houses
 5      */
 6     public int minCostII(int[][] costs) {
 7         if (costs == null || costs.length == 0 || costs[0].length == 0) return 0;
 8         
 9         for (int i = 1; i < costs.length; i++) {
10             for (int j = 0; j < costs[0].length; j++) {
11                 int tempMin = Integer.MAX_VALUE;
12                 for (int k = 0; k < costs[0].length; k++) {
13                     if (k != j) {
14                         if (tempMin > costs[i - 1][k]) {
15                             tempMin = costs[i - 1][k];
16                         }
17                     }
18                 }
19                 costs[i][j] += tempMin;
20             }
21         }
22         int tempMin = Integer.MAX_VALUE;
23         for (int j = 0; j < costs[0].length; j++) {
24             if (costs[costs.length - 1][j] < tempMin) {
25                 tempMin = costs[costs.length - 1][j];
26             }
27         }
28         return tempMin;
29     }
30 }

 

转载于:https://www.cnblogs.com/beiyeqingteng/p/5655104.html