There are a row of n houses, each house can be painted with one of the three colors: red, blue or green. 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 3 cost matrix. For example, costs[0][0] is the cost of painting house 0 with color red; costs[1][2] is the cost of painting house 1 with color green, and so on... Find the minimum cost to paint all houses.
Notice
All costs are positive integers.
Example
Given costs = [[14,2,11],[11,14,5],[14,3,10]] return 10
house 0 is blue, house 1 is green, house 2 is blue, 2 + 5 + 3 = 10
LeetCode上的原题，请参见我之前的博客Paint House。
解法一：
class Solution { public: /** * @param costs n x 3 cost matrix * @return an integer, the minimum cost to paint all houses */ int minCost(vector<vector<int>>& costs) { if (costs.empty() || costs[0].empty()) return 0; vector<vector<int>> dp = costs; for (int i = 1; i < dp.size(); ++i) { for (int j = 0; j < 3; ++j) { dp[i][j] += min(dp[i - 1][(j + 1) % 3], dp[i - 1][(j + 2) % 3]); } } return min(dp.back()[0], min(dp.back()[1], dp.back()[2])); } };
解法二：
class Solution { public: /** * @param costs n x 3 cost matrix * @return an integer, the minimum cost to paint all houses */ int minCost(vector<vector<int>>& costs) { if (costs.empty() || costs[0].empty()) return 0; vector<vector<int>> dp = costs; for (int i = 1; i < dp.size(); ++i) { dp[i][0] += min(dp[i - 1][1], dp[i - 1][2]); dp[i][1] += min(dp[i - 1][0], dp[i - 1][2]); dp[i][2] += min(dp[i - 1][0], dp[i - 1][1]); } return min(dp.back()[0], min(dp.back()[1], dp.back()[2])); } };
本文转自博客园Grandyang的博客，原文链接：粉刷房子[LintCode] Paint House ，如需转载请自行联系原博主。
