LeetCode刷题——Array篇(Best Time to Buy and Sell Stock II)

lpcarl

于 2018-04-16 16:22:48 发布

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分类专栏： Array

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本文介绍了一种算法，用于计算股票交易的最大利润。该算法允许进行多次买卖操作，但每次卖出后才能再次购买。通过分析价格波动，算法在价格上升时买入、下降时卖出，以实现利润最大化。

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Say you have an array for which the ith 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).

思路：
官网提供的一种是将所有递增的结果加起来；
这个有点像电路里的上升沿和下降沿，因此，在上升沿加入买入，下降沿售出才能使收益最大
代码如下

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int sum = 0;
        bool flag = false;
        int begin = 0;
        int end = 1;
        prices.push_back(0);
        for (int i = 0; i < prices.size(); i++)
        {
            if (prices[i + 1] > prices[i]&&flag==false)
            {
                flag = true;
                begin = i;
            }
            else if (prices[i + 1]< prices[i]&&flag==true)
            {
                flag = false;
                end = i;
                sum += prices[end] - prices[begin];
            }
        }
        return sum;

    }
};