265. Paint House II

问题描述：

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?

解题思路：

题目要求求最小的花费，我们会首先想到动态规划。

题目中要求的是相邻的两个房子的颜色不能相等, 若我们给当前第i个房子涂上颜色j, 则代表第i-1个房子不能涂颜色j

我们可以用数组lowest[j]表示涂到当前房屋除了涂j个颜色在其他选择中最小的花费。

则我们在计算下一个房子涂颜色j时的总花费为：lowest[j] + costs[i][j]

所以我们在每一次计算完这个房子所有可能的耗费后，要更新lowest为当前房子的情况。

需要注意的是，这样比较不能涵盖只有一个房子和一个颜色的情况，所以需要单独列出来。

代码：

class Solution {
public:
    int minCostII(vector<vector<int>>& costs) {
        int n = costs.size();
        if(n == 0) return 0;
        int m = costs[0].size();
        if(m == 0) return 0;
        
        if(m == 1 && n == 1) return costs[0][0];
        
        vector<int> lowest(m, 0);
        for(int i = 0; i < n; i++){
            vector<int> temp(m, INT_MAX);
            for(int j = 0; j < m; j++){
                lowest[j] += costs[i][j];
                if(j > 0){
                    temp[j] = min(temp[j-1], lowest[j-1]);
                }
            }
            int low = lowest[m-1];
            for(int j = m-2; j > -1; j--){
                temp[j] = min(temp[j], low);
                low = min(low, lowest[j]);
            }
            lowest = temp;
        }
        int ret = lowest[0];
        for(int i = 0; i < m; i++){
            ret = min(ret, lowest[i]);
        }
        return ret;
    }
};