【EASY】Paint House

# 问题

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 is the cost of painting house 0 with color red; costs 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] += min(dp[i - 1], dp[i - 1])
• dp[i] += min(dp[i - 1], dp[i - 1])
• dp[i] += min(dp[i - 1], dp[i - 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

``````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] += min(dp[i - 1], dp[i - 1]);
// 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] += min(dp[i - 1], dp[i - 1]);
// 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] += min(dp[i - 1], dp[i - 1]);
}

return min(min(dp.back(), dp.back()), dp.back());
}
};``````