# 问题 原题链接：[https://leetcode.com/problems/paint-house/](https://leetcode.com/problems/paint-house/) 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. Note: - All costs are positive integers. Example: ``` Input: [[17,2,17],[16,16,5],[14,3,19]] Output: 10 Explanation: Paint house 0 into blue, paint house 1 into green, paint house 2 into blue. Minimum cost: 2 + 5 + 3 = 10. ``` # 分析过程 - 输入：一个二维数组，N行，3 列，每一列代表用 RBG 中的一种颜色刷这个房子的 cost。 - 输出：刷完所有房子的 cost 的最小值。 - 思路：动态规划，使用临时空间，累计 cost。第 i 个房子用红色的 cost 等于这个房子用红色的 cost 加上第 i - 1 个房子用蓝色或绿色的累计 cost 中小的那个。 - 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]) - 结束以后最后一行就是累计的 cost，找到最小的即可。 # 解决方法 ```cpp class Solution { public: int minCost(vector<vector<int>>& costs) { if (costs.size() == 0) { return 0; } // 临时空间，或者可以在原基础上修改 // 存放涂第i个房间时，用该颜色的最小cost（累计） vector<vector<int>> dp = costs; for (int i = 1; i < dp.size(); ++i) { // 第i个房间用红色的cost等于用红色的cost加上上一个房间用蓝色或绿色中累计cost较小的那个 dp[i][0] += min(dp[i - 1][1], dp[i - 1][2]); // 第i个房间用绿色的cost等于用红色的cost加上上一个房间用蓝色或红色中累计cost较小的那个 dp[i][1] += min(dp[i - 1][0], dp[i - 1][2]); // 第i个房间用蓝色的cost等于用红色的cost加上上一个房间用绿色或红色中累计cost较小的那个 dp[i][2] += min(dp[i - 1][0], dp[i - 1][1]); } return min(min(dp.back()[0], dp.back()[1]), dp.back()[2]); } }; ```