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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.
Dynamic Programming
dp[i][j] = costs[i][j] + min(dp[i-1][(j+1)%3], dp[i-1][(j+2)%3])
需要维持dp[i] 和dp[i-1] 。
方法一：可以直接建立数组dp[costs.length][3]，花更多空间。
方法二：也可以建立两个preRow[3] 储存dp[i-1], currORw 储存dp[i], 节省空间，运行时间也短一些。
Java code:
方法一：
public class Solution { public int minCost(int[][] costs) { if(costs == null || costs.length == 0) { return 0; } int len = costs.length; int[][] dp = new int[len][3]; dp[0] = costs[0]; for(int i = 1; i < len; i++) { for(int j = 0; j < 3; j++) { dp[i][j] = costs[i][j] + Math.min(dp[i-1][(j+1)%3], dp[i-1][(j+2)%3]); } } return Math.min(Math.min(dp[len-1][0], dp[len-1][1]), dp[len-1][2]); } }
方法二：
public class Solution { public int minCost(int[][] costs) { if(costs == null || costs.length == 0){ return 0; } int[] preRow = costs[0]; int len = costs.length; for(int i = 1; i < len; i++) { int[] currRow = new int[3]; for(int j = 0; j < 3; j++){ currRow[j] = costs[i][j] + Math.min(preRow[(j+1)%3], preRow[(j+2)%3]); } preRow = currRow; } return Math.min(Math.min(preRow[0], preRow[1]), preRow[2]); } }
Reference:
1. https://leetcode.com/discuss/63863/simple-15-line-code-with-o-n-time-and-o-1-memory-solution-java
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原文地址：http://www.cnblogs.com/anne-vista/p/4894317.html
