[LeetCode265]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.

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

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這題這麼不簡單，我當然不會。參考了大家的O(nk)方法。
dp[j]: 刷前i個房子且最後一個房子用的是j顏色的minimum cost。
min1: 刷前i個房子最後用某個顏色時最小的cost。
min2: 刷前i個房子最後用另外某個顏色時第二小的cost。

對於當前房子，如果之前的房子的minimum cost並不是min1 我們可以直接選取min1的顏色，否則選取min2的顏色。因爲只要相鄰兩個房子顏色不一樣即可。

class Solution {
public:
    int minCostII(vector<vector<int>>& costs) {
        if (costs.empty()) return 0;
        int n = costs.size(), k = costs[0].size(), min1 = 0, min2 = 0;
        vector<int> dp(k,0);
        for(int i = 0; i<n; ++i){
            int tmp1 = min1, tmp2 = min2;
            min1 = min2 = INT_MAX;
            for(int j = 0; j<k; ++j){
                dp[j] = ((dp[j] == tmp1) ? tmp2 : tmp1) + costs[i][j];
                if(min1 <= dp[j]) min2 = min(min2, dp[j]);
                else{
                    min2 = min1;
                    min1 = dp[j];
                }
            }
        }
        return min1;
    }
};