256. Paint House

问题描述：

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.

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.

 

解题思路：

这里给出了红绿蓝三种颜色来涂房子，要求我们用最小的耗费来做。

因为颜色个数就只有3个，所以我们可以直接取另外两个最小的值。

当前记录的是，到这个房子位置，涂该颜色最小的耗费

 

代码：

class Solution {
public:
    int minCost(vector<vector<int>>& costs) {
        int n = costs.size();
        if(n == 0) return 0;
        for(int i = 1; i < n; i++){
            costs[i][0] += min(costs[i-1][1], costs[i-1][2]);
            costs[i][1] += min(costs[i-1][0], costs[i-1][2]);
            costs[i][2] += min(costs[i-1][0], costs[i-1][1]);
        }
        
        return min(costs[n-1][0], min(costs[n-1][1], costs[n-1][2]));
    }
};