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.

All costs are positive integers.Have you met this question in a real interview? YesProblem Correction

Example

Given n = 3, k = 3, costs = [[14,2,11],[11,14,5],[14,3,10]] return 10

house 0 is color 2, house 1 is color 3, house 2 is color 2, 2 + 5 + 3 = 10

Challenge

Could you solve it in O(nk)?

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：

public class Solution {
    /**
     * @param costs: n x k cost matrix
     * @return: an integer, the minimum cost to paint all houses
     */
    public int minCostII(int[][] costs) {
        // write your code here
        
        if(costs == null || costs.length == 0 || costs[0].length == 0)
            return 0;
            
            
        int[][] dp = new int[costs.length][costs[0].length];
        
        int min1 = -1, min2 = -1;
        
        for (int i = 0; i < costs.length; i++ ){
            int last1 = min1, last2 = min2;
            
            min1 = -1;
            min2 = -1;
            
            for(int j = 0; j < costs[0].length; j++){
                
                if(j != last1){
                    dp[i][j] = costs[i][j] + (last1 < 0 ? 0 : dp[i-1][last1]);
                }
                
                else{
                    dp[i][j] = costs[i][j] + (last2 < 0 ? 0 : dp[i-1][last2]);
                }
                
                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[costs.length-1][min1];
        
        
    }
}