problem 122. Best Time to Buy and Sell Stock II 这道题由于可以无限次买入和卖出。我们都知道炒股想挣钱当然是低价买入高价抛出,那么这里我们只需要从第二天开始,如果当前价格比之前价格高,则把差值加入利润中,因为我们可以昨天买入,今日卖出,若明日价更高的话,还可以今日买入,明日再抛出。以此
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...
leetcode-cn.com/problem 记录每个元素之前的最小值,便利一遍就行 122买卖股票最佳时机2 leetcode-cn.com/problem 便利一次数组,发现前面大于后面就累加,代表吃到了所有利润 124二叉树最大路径和 leetcode-cn.com/problem 在普通求深度的基础上,考虑两种情况root +Max (left,right)和 root+left+right ,只有第一...
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...
Solution to the problem Unit test Note that each of these problemshave passedtheir respective test cases on LeetCode. The unit tests included with each solution in this repo are not comprehensive. They serve as a quick way to drop in a test case, hook up the debugger, and step through th...
Leetcode solution 772: Basic Calculator III Problem Statement Implement a basic calculator to evaluate a simple expression string. The expression string may contain open(and closing parentheses), the plus+or minus sign-,non-negative integers and empty spaces....
*/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: 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()]){ ...
The solution set must not contain duplicate subsets. Day 1670 答案揭晓 DS Interview Question & Answer What is root cause analysis? Root cause analysis (RCA) is a method of problem-solving used for identifying the root causes of faults or problems. A factor is considered a root cause of remo...