LeetCode 122. Best Time to Buy and Sell Stock II 题解（C++）

LeetCode 122. Best Time to Buy and Sell Stock II 题解（C++）

题目描述

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

使用贪心算法。这里我们从第一个元素开始遍历，每次若该第i个元素小于第i+1个元素（即第i+1天的股票价格高于第i天的股票价格），则卖出该股票；

若第i+1个元素仍小于第i+2个元素（即第i+2天的股票价格高于第i+1天的股票价格），则我们再次买进第i+1天的股票，并在第i+2天卖出；比如prices序列的值为3,4,5,6，则最高收益等于6-3 = （4-3）+（5-4）+（6-5）；

若第i+1个元素大于第i+2个元素，则不买进股票。

这里贪心算法是指我们只考虑最近的利益，若今天可以得到利益，就卖出股票，而不需要考虑以后是否有更高的卖出价格。

代码

class Solution
{
public:
int maxProfit(vector<int>& prices)
{
int profit = 0;
for (int i = 1; i < prices.size(); ++i)
{
if (prices[i - 1] < prices[i])
{
profit += prices[i] - prices[i - 1];
}
}

return profit;
}
};