• 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:

    https://leetcode.com/discuss/51721/simple-java-dp-solution 

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  • 原文地址:https://www.cnblogs.com/yrbbest/p/5014958.html
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