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.
[分析]
动态规划题目,我习惯于从目的倒推中间过程,最后一家house选用什么颜色能使总花费最小呢,假设知道前面第 N - 1家使用不同颜色时的总花费,那问题就解决了,因此中间计算需要存储的信息dp[i][j] 表示第 i 家选用 颜色 j 时 0 - i家house所需的最小花费,递推式见代码。
public class Solution { public int minCost(int[][] costs) { if (costs == null || costs.length == 0 || costs[0].length == 0) return 0; int N = costs.length; int[][] dp = new int[2][3]; int prev = 0, curr = 1; for (int i = 0; i < N; i++) { prev = curr; curr = 1 - curr; dp[curr][0] = Math.min(dp[prev][1], dp[prev][2]) + costs[i][0]; dp[curr][1] = Math.min(dp[prev][0], dp[prev][2]) + costs[i][1]; dp[curr][2] = Math.min(dp[prev][0], dp[prev][1]) + costs[i][2]; } return Math.min(dp[curr][0], Math.min(dp[curr][1], dp[curr][2])); } }