256. Paint House
题目:
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.
链接: http://leetcode.com/problems/paint-house/
题解:
房子用RGB刷漆, 每种漆对于每栋房子来说花费不一样,要求相邻两房子不一个色,并且花费最小。 这道题看提示需要用dp做。一开始我想设一个sum,一个last color = -1,不过没办法解决duplicate的问题。所以还是参考了jianchao.li大神的代码,对RGB分别进行dp。
Time Complexity - O(n), Space Complexity - O(1)
public class Solution { public int minCost(int[][] costs) { //dp if(costs == null || costs.length == 0) { return 0; } int len = costs.length, red = 0, blue = 0, green = 0; for(int i = 0; i < costs.length; i++) { int prevRed = red, prevBlue = blue, prevGreen = green; red = costs[i][0] + Math.min(prevBlue, prevGreen); blue = costs[i][1] + Math.min(prevRed, prevGreen); green = costs[i][2] + Math.min(prevRed, prevBlue); } return Math.min(red, Math.min(blue, green)); } }
二刷:
继续锻炼思维能力,现在写这类题目已经比较轻松了,看来自己真的是在进步的。
Java:
Time Complexity - O(n), Space Complexity - O(1)
public class Solution { public int minCost(int[][] costs) { if (costs == null || costs.length == 0) return 0; int pRed = 0, pGreen = 0, pBlue = 0; for (int[] cost : costs) { int lastRed = pRed, lastGreen = pGreen, lastBlue = pBlue; pRed = cost[0] + Math.min(lastGreen, lastBlue); pGreen = cost[1] + Math.min(lastRed, lastBlue); pBlue = cost[2] + Math.min(lastRed, lastGreen); } return Math.min(pRed, Math.min(pGreen, pBlue)); } }
Reference: