[Swift]LeetCode265.粉刷房子 II $ Paint House II

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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.

Follow up:
Could you solve it in O(nk) runtime?

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 有一排N栋房子，每栋房子均可以涂上其中一种K颜色。用某种颜色粉刷每栋房子的费用是不一样的。你必须把全部的房子都漆成没有两个相邻的房子有相同的颜色。

用一个n x k的成本矩阵表示每栋房子涂上某种颜色的成本。例如，costs[0][0]是用颜色0绘制房子0的成本；costs[1][2]是用颜色2绘制房子1的成本，等等…找出油漆全部房屋的最低成本。

注：

全部成本都是正整数。

进阶：

你能在运行时解决它吗？

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 1 class Solution {
 2     func minCostII(_ costs: [[Int]]) -> Int {
 3         if costs.isEmpty || costs[0].isEmpty
 4         {
 5             return 0
 6         }
 7         var min1:Int = 0
 8         var min2:Int = 0
 9         var idx1:Int = -1
10         for i in 0..<costs.count
11         {
12             var m1:Int = Int.max
13             var m2:Int = m1
14             var id1:Int = -1
15             for j in 0..<costs[0].count
16             {
17                 var cost:Int = costs[i][j] + (j == idx1 ? min2 : min1)
18                 if cost < m1
19                 {
20                     m2 = m1
21                     m1 = cost
22                     id1 = j
23                 }
24                 else if cost < m2
25                 {
26                      m2 = cost
27                 }
28             }
29             min1 = m1
30             min2 = m2
31             idx1 = id1
32         }
33         return min1
34     }
35 }