题目描述

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.

样例

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.

算法1

(DP) $O(n)$

dp[i][j]表示第i行第j列的最小cost。由于相邻两行不能在同一列选择，所以dp[i][j]来自dp[i-1][j+1]或者dp[i-1][j+2]。考虑到j+1和j+2不能越界，取模3即可。

时间复杂度

O（n）
空间复杂度是O（n*3）

参考文献

C++ 代码

class Solution {
public:
    int minCost(vector<vector<int>>& costs) {
        int n=costs.size();
        if(n==0) return 0;
        vector<vector<int>> dp(costs);

        for(int i=1;i<n;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[n-1][0], dp[n-1][1],dp[n-1][2]});
    }
};