LeetCode122:Best Time to Buy and Sell Stock II-优快云博客

本文介绍了一种算法，用于计算给定股票价格数组时的最大交易利润。算法允许进行多次买入和卖出操作，但不能同时持有多个股票。通过一次遍历数组，比较相邻两天的价格，当后一天价格高于前一天时，累加差值得到最大利润。

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题目：

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).

解题思路：

这题与上一题Best Time to Buy and Sell Stock相比，其实还更简单，扫描一遍数组，当前前元素大于前一个元素，则将其差值累加到max中，最后求得的max即为最大利益。

代码：

#include <iostream>
#include <vector>

using namespace std;

/**
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).

*/

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

int main(void)
{
    int arr[] = {2,4,5,1,7,10};
    int n = sizeof(arr) / sizeof(arr[0]);
    vector<int> stock(arr, arr+n);
    Solution solution;
    int max = solution.maxProfit(stock);
    cout<<max<<endl;
    return 0;
    return 0;
}