[LeetCode] Paint House Problem Description: There are a row ofnhouses, 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...
[LeetCode] 256. Paint House 粉刷房子 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 ...
ans != MAX: ans 即为满足题意的所有染色方案中,花费最小的费用 DP 常见的三种优化方式见LeetCode 583 这题的思路。 本题可以采用滚动数组的方式进行优化,因为每一行的状态依赖上一行的所有状态,所以无法采用其他两种方式进行优化。 使用滚动数组的话,能将空间复杂度从O(m * target * n) 优化为 O(target *...
There are a row of houses, each house can be painted with three colors red, blue and 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. You have to paint the houses with mi...
Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5. Follow up: Could you solve it in O(nk) runtime? 答案 直接改编Paint House题的答案 O(nk^2) classSolution{publicintminCostII(int[][]costs){if(costs.length==0)return0;intn=costs.length,k=cost...
Explanation: Paint house 0 into blue, paint house 1 into green, paint house 2 into blue. Minimum cost: 2 + 5 + 3 = 10. 这道题说有n个房子,每个房子可以用红绿蓝三种颜色刷,每个房子的用每种颜色刷的花费都不同,限制条件是相邻的房子不能用相同的颜色来刷,现在让求刷完所有的房子的最低花费是...
[LeetCode] 256. Paint House There is a row ofnhouses, where each house can be painted one of 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...
今天坐在Victoria coffee 里面从第一个最简单的paint house问题解决开始。 ps:这里都是博主自己做题时的思考过程,不是大神讲解,欢迎指正缺陷和错误。 Paint House 题目链接:https://leetcode.com/problems/paint-house/There are a row of n houses, each house can be painted with one of the three colors:...
LeetCode "Paint House" A typical DP class Solution { public: int minCost(vector<vector<int>>& costs) { size_t len = costs.size(); if(len == 0) return 0; vector<vector<int>> dp(len, vector<int>(3, 0)); dp[0] = costs[0]; for(int i = 1; i < len; i ++) { for(...
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. ...