There are a row of n houses, each house can be painted with one of the three colors: red, blue or green. 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 3 cost matrix. For example, costs[0][0] is the cost of painting house 0 with color red; costs[1][2] is the cost of painting house 1 with color green, and so on... Find the minimum cost to paint all houses.
Note:All costs are positive integers.
Example:
Input: [[17,2,17],[16,16,5],[14,3,19]] Output: 10 Explanation: Paint house 0 into blue, paint house 1 into green, paint house 2 into blue. Minimum cost: 2 + 5 + 3 = 10.
题意：
一排有n套房子，每个房子可以选一种颜色（红色、蓝色或者绿色）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。
Solution1： DP
scan一遍n套房子：
costs[i][0] = costs[i][0] + min(costs[i-1][1], costs[i-1][2])
当下选0号色的花费(和) = 当下选0号色的花费(单价) + 前套房子选1号色或者选2号色中较小花费
code
1 class Solution { 2 public int minCost(int[][] costs) { 3 if(costs == null || costs.length == 0 || costs[0].length < 3) return 0; 4 5 for(int i = 1; i < costs.length; i++ ){ // 照顾[i-1]所以 i 从 1 开始， 则 i - 1 从 0 开始。若从0开始，i 就只能取到costs.length-1 6 costs[i][0] += Math.min(costs[i-1][1], costs[i-1][2]); 7 costs[i][1] += Math.min(costs[i-1][0], costs[i-1][2]); 8 costs[i][2] += Math.min(costs[i-1][1], costs[i-1][0]); 9 } 10 int n = costs.length-1; 11 return Math.min(Math.min(costs[n][0], costs[n][1]), costs[n][2]); 12 } 13 }
转载于:https://www.cnblogs.com/liuliu5151/p/9108844.html
