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: