[LeetCode] 265. Paint House II 粉刷房子之二
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.
Follow up:
Could you solve it in O(nk) runtime?
这道题是之前那道 Paint House 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,那么用 min2 对应的值计算,反之用 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<vector<int>>& costs) { if (costs.empty() || costs[0].empty()) return 0; vector<vector<int>> 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<vector<int>>& 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; } };
Github 同步地址:
https://github.com/grandyang/leetcode/issues/265
类似题目:
参考资料:
https://leetcode.com/problems/paint-house-ii/
https://leetcode.com/problems/paint-house-ii/discuss/69509/Easiest-O(1)-space-JAVA-solution
https://leetcode.com/problems/paint-house-ii/discuss/69492/AC-Java-solution-without-extra-space