265. Paint House II

There are a row of n houses, each house can be painted with one of the k colors. 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 k cost matrix. For example, costs[0][0] is the cost of painting house 0 with color 0; costs[1][2] is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.

Note:
All costs are positive integers.

Example:

Input: [[1,5,3],[2,9,4]]
Output: 5
Explanation: Paint house 0 into color 0, paint house 1 into color 2. Minimum cost: 1 + 4 = 5; 
             Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5. 

Follow up:
Could you solve it in O(nk) runtime?

A:
DP
每次记录之前一行，最小的value和index，以及 secondMinimum value;

class Solution {
public:
    int minCostII(vector<vector<int>>& costs) {
        int n = costs.size();
        if(n==0)
            return 0;
        int k = costs[0].size();
        int minVal = 0, minIndex=0;
        int min2Val = 0; // no need to record its index
        for(auto & cost : costs){
            int newMinVal = INT_MAX-1, newMinIndex=0;
            int newMin2Val = INT_MAX;
            for(int i =0;i<k;i++){
                cost[i] += (i != minIndex ? minVal : min2Val);
                if(cost[i] < newMin2Val){
                    newMin2Val = cost[i];
                }
                if(cost[i] < newMinVal){
                    newMin2Val = newMinVal;
                    newMinVal = cost[i];
                    newMinIndex = i;
                }
            }
            minVal = newMinVal;
            minIndex = newMinIndex;
            min2Val = newMin2Val;
        }
        return minVal;
    }
};

不用单独把 第0行单独计算（只需要把minValue = 0 即可）

Mistakes: