Tuesday, December 29, 2015

256. Paint House ---------E

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.
A:
DP

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] ));
    }
};


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