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.

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

Solution: DP

这道题是之前那道Paint House的拓展，那道题只让用红绿蓝三种颜色来粉刷房子，而这道题让我们用k种颜色，这道题不能用之前那题的解法，会TLE。这题的解法的思路还是用DP，但是在找不同颜色的最小值不是遍历所有不同颜色，而是用min1和min2来记录之前房子的最小和第二小的花费的颜色，如果当前房子颜色和min1相同，那么我们用min2对应的值计算，反之我们用min1对应的值，这种解法实际上也包含了求次小值的方法，感觉也是一种很棒的解题思路，参见代码如下：

version 1: 16ms

class Solution {
public:
    int minCostII(vector<vector<int>>& costs) {
        if (costs.empty() || costs[0].empty()) return 0;
        int n = costs.size(), k = costs[0].size();
        int min1 = -1, min2 = -1;
        vector<vector<int>> dp = costs;
        for (int i = 0; i < n; ++i) {
            int last1 = min1, last2 = min2;
            min1 = min2 = -1;
            for (int j = 0; j < k; ++j) {
                if (j != last1) {
                    dp[i][j] += last1 < 0 ? 0:dp[i-1][last1];
                } else {
                    dp[i][j] += last2 < 0 ? 0:dp[i-1][last2];
                }
                // update min1 and min2
                if (min1 < 0 || dp[i][j] < dp[i][min1]) {
                    min2 = min1;
                    min1 = j;
                } else if (min2 < 0 || dp[i][j] < dp[i][min2]) {
                    min2 = j;
                }
            }

        }
        return dp.back()[min1];
    }
};

version 2: 16ms

下面这种解法不需要建立二维dp数组，直接用三个变量就可以保存需要的信息即可，参见代码如下：

class Solution {
public:
    int minCostII(vector<vector<int>>& costs) {
        if (costs.empty() || costs[0].empty()) return 0;
        int n = costs.size(), k = costs[0].size();
        int min1 = 0, min2 = 0, idx = -1;
        for (int i = 0; i < n; ++i) {
            int m1 = INT_MAX, m2 = INT_MAX, id1 = -1;
            for (int j = 0; j < k; ++j) {
                int cost = costs[i][j]+(j == idx ? min2:min1);
                if (cost < m1) {
                    m2 = m1;
                    m1 = cost;
                    id1 = j;
                } else if (cost < m2) {
                    m2 = cost;
                }
            }
            min1 = m1, min2 = m2, idx = id1;
        }
        return min1;
    }
};