[leetcode]265. Paint House II粉刷房子(K色可选)

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. 

 

题意：

一排有n套房子，每个房子可以选一种颜色（K种不同颜色）来粉刷。由于每个房子涂某种颜色的花费不同，求相邻房子不同色的粉刷最小花费。

做这个题的时候，想起跟老妈一起去过的Santa Cruz海棠花节的时候去过的这个网红海滩。

所以这个题，对于景区想打造网红般的colorful houses来讲，还是蛮有显示意义。

 

Solution1: DP

 

 

 

 

code

class Solution {
    public int minCostII(int[][] costs) {
        // sanity check
        if (costs == null || costs.length == 0) return 0;
        
        //init NO.1 house's curMin1, curMin2
        int curMin1 = -1; // most minimum
        int curMin2 = -1; // second minimum
        for (int j = 0; j < costs[0].length; j++) {
            if (curMin1 < 0 || costs[0][j] < costs[0][curMin1]) {
                curMin2 = curMin1;
                curMin1 = j;
            } else if (curMin2 < 0 || costs[0][j] < costs[0][curMin2]) {
                curMin2 = j;
            }
        }
        // scan from NO.2 house 
        for (int i = 1; i < costs.length; i++) {  // scan n houses
            // get NO.1 house's curMin1, curMin2
            int preMin1 = curMin1;
            int preMin2 = curMin2;
            // init NO.2 house's curMin1, curMin2
            curMin1 = -1;
            curMin2 = -1;
            // try k colors
            for (int j = 0; j < costs[0].length; j++) {
                if (j != preMin1) {  
                    costs[i][j] += costs[i - 1][preMin1];
                } else {
                    costs[i][j] += costs[i - 1][preMin2];
                }
                // update NO.2 house's curMin1, curMin2
                if (curMin1 < 0 || costs[i][j] < costs[i][curMin1]) {
                    curMin2 = curMin1;
                    curMin1 = j;
                } else if (curMin2 < 0 || costs[i][j] < costs[i][curMin2]) {
                    curMin2 = j;
                }
            }
        }
        return costs[costs.length - 1][curMin1];
    }
}