一.题目
Total Accepted: 68509 Total Submissions: 170952 Difficulty: Medium
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).
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二.解题技巧
这道题可以用贪心法来做,由于不限制次数,可以每一天都进行交易,然后将所有收入为正的部分都加起来,就是所获得的最大收益。
三.实现代码
#include <vector>
using std::vector;
class Solution
{
public:
int maxProfit(vector<int>& prices)
{
if (prices.empty())
{
return 0;
}
if (prices.size() == 1)
{
return 0;
}
int Result = 0;
int Tmp = 0;
for (vector<int>::iterator Ite = prices.begin(); Ite != prices.end() - 1;
++Ite)
{
Tmp = *(Ite + 1) - *Ite;
if (Tmp > 0)
{
Result += Tmp;
}
}
return Result;
}
};
四.体会
这道题用贪心法来解,非常简单,但是要想到贪心法却不简单。
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