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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?
有一排N栋房子，每栋房子都可以涂上其中一种K颜色。用某种颜色粉刷每栋房子的费用是不同的。你必须把所有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…找出油漆所有房屋的最低成本。
注：
所有成本都是正整数。
进阶：
你能在运行时解决它吗？
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 }