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Paint House 解答

时间:2015-11-09 07:05:49      阅读:238      评论:0      收藏:0      [点我收藏+]

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Question

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.

Solution

拿到这题的第一反应是画出解空间树。我们用1, 2, 3分别代表red, green, blue

      ()

     /  |  \

    1    2    3

          / \   / \   / \

   2  3 1  3 1  2 

   ...................

粗暴的方法是用DFS遍历整个解空间树。但是我们可以看到在每一层,其实有重复计算。

所以这题的思路和那道经典的求min path sum一样,是用DP。Time complexity O(n), space cost O(1)

cost1, cost2, cost3表示第n层选了1/2/3后的最少话费。

举个例子:

    red  green  blue

h1   1    2    3

h2   3    1    2

h3   4    3    2

我们从底向上遍历做DP

对于h3这一层:

cost1 = 4, cost2 = 3, cost3 = 2

对于h2这一层:

cost1‘ = 3 + min(cost2, cost3) = 5, cost2‘ = 1 + min(cost1, cost3) = 3, cost3‘ = 2 + min(cost1, cost2) = 5

对于h1这一层:

cost1‘‘ = 1 + min(cost2‘, cost3‘) = 4, cost2‘‘ = 2 + min(cost1‘, cost3‘) = 7, cost3‘‘ = 3 + min(cost1‘, cost2‘) = 6

因此最少话费是cost1‘‘

 1 public class Solution {
 2     public int minCost(int[][] costs) {
 3         if (costs == null || costs.length < 1) {
 4             return 0;
 5         }
 6         int m = costs.length, n = costs[0].length;
 7         int cost1 = costs[m - 1][0], cost2 = costs[m - 1][1], cost3 = costs[m - 1][2];
 8         for (int i = m - 2; i >= 0; i--) {
 9             int tmp1 = cost1, tmp2 = cost2, tmp3 = cost3;
10             cost1 = costs[i][0] + Math.min(tmp2, tmp3);
11             cost2 = costs[i][1] + Math.min(tmp1, tmp3);
12             cost3 = costs[i][2] + Math.min(tmp1, tmp2);
13         }
14         int result = Math.min(cost1, cost2);
15         return Math.min(result, cost3);
16     }
17 }

 

Paint House 解答

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原文地址:http://www.cnblogs.com/ireneyanglan/p/4948892.html

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