LeetCode 题解 | 121. 买卖股票的最佳时机 力扣(LeetCode) 【LeetCode】贪心算法--买卖股票的最佳时机 II(122) 一、写在前面 为什么要在LeetCode刷题?大家都知道不管是校招还是社招算法题是必考题,而这一部分恰巧是大多数人的短板,所以刷题首先是为了提高自身的编程能力,能够在算法面试中脱颖而出… 玩数据的...
position[i] + 1orposition[i] - 1withcost = 1. Returnthe minimum costneeded to move all the chips to the same position. Example 1: Input:position = [2,2,2,3,3]Output:2Explanation:We can move the two chips at position3toposition2.Eachmove has cost =1. The total cost =2. Exampl...
原题链接在这里:https://leetcode.com/problems/minimum-cost-to-merge-stones/ 题目: There areNpiles of stones arranged in a row. Thei-th pile hasstones[i]stones. Amoveconsists of merging exactlyKconsecutive piles into one pile, and the cost of this move is equal to the total number of s...
LeetCode 1595问题中动态规划的状态转移方程是什么? 题目 题解: 动态规划,用二进制压缩状态,注意分析几种情况,就能推出来正确的状态转移方程。 代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution { public: int dp[12][4096]; int connectTwoGroups(vector<vector<int>>& cost) { int n =...
Runtime: 21 ms, faster than 100.00% of Java online submissions for Minimum Cost to Convert String I. Memory Usage: 45.7 MB, less than 33.33% of Java online submissions for Minimum Cost to Convert String I. --- 最主要的就是这个转换的函数,对于每个位置,从i到j都要去找一下是否有更低的成...
Can you solve this real interview question? Minimum Cost to Cut a Stick - Given a wooden stick of length n units. The stick is labelled from 0 to n. For example, a stick of length 6 is labelled as follows: [https://assets.leetcode.com/uploads/2020/07/21
力扣LeetCode中文版,码不停题 -全球极客编程职业成长社区 🎁 每日任务|力扣 App|百万题解|企业题库|全球周赛|轻松同步,使用已有积分换礼 × Testcase Test Result Test Result You have planned some train traveling one year in advance. The days of the year in which you will travel are given as an...
题目地址:https://leetcode-cn.com/problems/connecting-cities-with-minimum-cost/ 题目描述 There areNcities numbered from1toN. You are given connections, where eachconnections[i] = [city1, city2, cost]represents the cost to connectcity1andcity2together. (A connection is bidirectional: connectingci...
一道常规求minimum spanning tree的题,返回值为最小cost,出现在热带雨林厂的OA中。 难度:medium 算法:Kruskal,并查集 思路: 求MST用Kruskal或者Prim,理论上Prim在非常dense的graph中更快一点,否则Kruskal更快也更简单。 这里采用Kruskal,先对边排序,每次连最小边,此过程要记录component的数量,利用并查集实现树的合并(...
cost(k) 表示 从k处切断的cost 代码语言:javascript 代码运行次数:0 classSolution{public:int dp[105][105];intgetLen(int i,int j,vector<int>&cuts,int n){int left=i==0?0:cuts[i-1];int right=j==cuts.size()-1?n:cuts[j+1];returnright-left;}intminCost(int n,vector<int>&cuts){...