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?这道题是之前那道的拓展，那道题只让用红绿蓝三种颜色来粉刷房子，而这道题让我们用k种颜色，这道题不能用之前那题的解法，会TLE。这题的解法的思路还是用DP，但是在找不同颜色的最小值不是遍历所有不同颜色，而是用min1和min2来记录之前房子的最小和第二小的花费的颜色，如果当前房子颜色和min1相同，那么我们用min2对应的值计算，反之我们用min1对应的值，这种解法实际上也包含了求次小值的方法，感觉也是一种很棒的解题思路，参见代码如下：
解法一：
class Solution {public: int minCostII(vector >& costs) { if (costs.empty() || costs[0].empty()) return 0; vector > dp = costs; int min1 = -1, min2 = -1; for (int i = 0; i < dp.size(); ++i) { int last1 = min1, last2 = min2; min1 = -1; min2 = -1; for (int j = 0; j < dp[i].size(); ++j) { if (j != last1) { dp[i][j] += last1 < 0 ? 0 : dp[i - 1][last1]; } else { dp[i][j] += last2 < 0 ? 0 : dp[i - 1][last2]; } if (min1 < 0 || dp[i][j] < dp[i][min1]) { min2 = min1; min1 = j; } else if (min2 < 0 || dp[i][j] < dp[i][min2]) { min2 = j; } } } return dp.back()[min1]; }}; 下面这种解法不需要建立二维dp数组，直接用三个变量就可以保存需要的信息即可，参见代码如下：
解法二：
class Solution {public: int minCostII(vector >& costs) { if (costs.empty() || costs[0].empty()) return 0; int min1 = 0, min2 = 0, idx1 = -1; for (int i = 0; i < costs.size(); ++i) { int m1 = INT_MAX, m2 = m1, id1 = -1; for (int j = 0; j < costs[i].size(); ++j) { int cost = costs[i][j] + (j == idx1 ? min2 : min1); if (cost < m1) { m2 = m1; m1 = cost; id1 = j; } else if (cost < m2) { m2 = cost; } } min1 = m1; min2 = m2; idx1 = id1; } return min1; }}; 本文转自博客园Grandyang的博客，原文链接：，如需转载请自行联系原博主。