[LeetCode] Best Time to Buy and Sell Stock II Solution

最新推荐文章于 2025-04-15 23:58:56 发布

原创最新推荐文章于 2025-04-15 23:58:56 发布 · 175 阅读

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本文介绍了一种解决无限次股票交易问题的算法，通过扫描股价数组并计算所有正价差之和来确定最大利润。该方法允许在进行多次买卖操作的同时避免同时持有股票。

Say you have an array for which the i ^th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

» Solve this problem

[Thoughts]
A bit different with previous one. Since we can make unlimited transactions, this question turns to sum all the positive price difference.

So, scan from left to right, and add all positive diff value.

[Code]

1:    int maxProfit(vector<int> &prices) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      int max=0;  
5:      int sum = 0;  
6:      for(int i =1; i< prices.size(); i++)  
7:      {  
8:        int diff = prices[i] -prices[i-1];  
9:        if(diff>0)  
10:          sum+=diff;  
11:      }  
12:      return sum;  
13:    }