Problem:https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/ Say you have an array for which theith element is the price of a given stock on dayi. Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell ...
// Solution 1: This problem can be quite challenging if it's your first attempt. I didn't work it out after almost two hour, so I referred to this solution from the internet. Very clever and efficient. 代码1 //Code 1 117 Populating Next Right Pointers in Each Node II // #117 填充...
84. 柱状图中最大的矩形对每个位置求一个最大的面积,即找到左边第一个比自己小的位置和右边第一个比自己小的位置,求一下面积,使用单调栈即可 https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/solution/dong-hua-yan-shi-san-chong-shi-xian-116-tian-chong/ 121 买卖股票最佳...
Problemset LeetCode Record Oct 3, 2020 README.md LeetCode Record Oct 3, 2020 README_EN.md LeetCode Record Oct 3, 2020 Repository files navigation README | English | 简体中文 |LeetCode 的解答最近一次更新: 2020-10-03 18:52:32 The...
My solution to this problem is by DFS and backtracking. The difference is that I added some optimization strategies to the program: 1. [7][6] can be represented as [7 * 9 + 6], you don't have to calculate it every time, store the coordinates in an array instead. ...
122 29.4K 47zbjoy ・ 2025.04.20 通过0来判断是否可以到达终点(新思路) Problem: 思路 我找到了一个好玩的方法来解决这个问题:\n在跳跃的过程中,\n\n只要没有0的话,显然一定可以到终点\n \n如果有0: 终点的0: 没有影响,可以忽略 非终点的0:只有当前面 C++ 1 11 0立志...
(Notes: 🔒 means you need to buy a book from Leetcode to unlock the problem)#TitleSource CodeArticleDifficulty 1 two-sum Python Javascript 📝 Easy 2 add-two-numbers Python 📝 Medium 3 longest-substring-without-repeating-characters Python 📝 Medium 4 median-of-two-sorted-arrays Python ...
*/classSolution{privateintgetNext(int n){int totalSum=0;while(n>0){int d=n%10;n=n/10;totalSum+=d*d;}returntotalSum;}publicbooleanisHappy(int n){Set<Integer>seen=newHashSet<>();while(n!=1&&!seen.contains(n)){seen.add(n);n=getNext(n);}returnn==1;}} ...
class Solution { public: bool combine(int min, int k, int n, vector<int> vec, vector<vector<int>>& result){ if(n<min*k) //更确切的。能够用等差数列公式算出最大值最小值 return false; if(n>9*k) return true; if(k==1)
class Solution { public: int trap(vector<int>& height) { if(height.empty()) return 0; height.push_back(0); int res =0; stack<int> st;//decreasing stack for(int i = 0; i<height.size();++i){ while(!st.empty() && height[i] > height[st.top()]){ ...