[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。

思路：

维护 lastMinCost1（前一套房子的最小花费颜色色号）和 lastMinCost2（前一套房子次小花费颜色色号）

若当前房子颜色 j 色号等于前一套房子的最小花费颜色色号lastMinCost1，则目前为止粉刷房子的最小花费costs[i][j] = costs[i][j] + costs[i-1][lastMinCost2]

否则，若当前房子颜色 j 色号等于前一套房子的次小花费颜色色号lastMinCost2，则目前为止粉刷房子的最小花费costs[i][j] = costs[i][j] + costs[i-1][lastMinCost1]

代码：

 1   public int minCostII(int[][] costs) {
 2         // sanity check
 3         if (costs == null || costs.length == 0) return 0;
 4 
 5         int n = costs.length;  // n houses
 6         int k = costs[0].length; // k colors
 7         int minCost1 = -1; // most minimum
 8         int minCost2 = -1; // second minimum
 9         for (int i = 0; i < n; i++) {  // scan n houses
10 
11             int lastMinCost1 = minCost1; // 之前最小花费的颜色色号，先存下来
12             int lastMinCost2 = minCost2; // 之前次小花费的颜色色号，先存下来
13 
14             minCost1 = -1;
15             minCost2 = -1;
16 
17             // scan k colors
18             for (int j = 0; j < k; j++) {
19                 if (j != lastMinCost1) {  // 若当前颜色的色号不等于最小花费的颜色色号，就加上最小的开销
20                     costs[i][j] += lastMinCost1 < 0 ? 0 : costs[i - 1][lastMinCost1];
21                 } else { // 若当前颜色的色号等于最小花费的颜色色号，就加上次小的开销
22                     costs[i][j] += lastMinCost2 < 0 ? 0 : costs[i - 1][lastMinCost2];
23                 }
24                 //更新当前最小和次小开销
25                 if (minCost1 < 0 || costs[i][j] < costs[i][minCost1]) {
26                     minCost2 = minCost1;
27                     minCost1 = j;
28                 } else if (minCost2 < 0 || costs[i][j] < costs[i][minCost2]) {
29                     minCost2 = j;
30                 }
31             }
32         }
33         return costs[n - 1][minCost1];
34     }

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

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