关于股票最佳买卖时机的lintcode代码

Posted 2020-09-03

class Solution {
public:
    /**
     * @param prices: Given an integer array
     * @return: Maximum profit
     */
    int maxProfit(vector<int> &prices) {
        // write your code here
if(prices.empty())
return 0;
int maxpro=0;
int minprofit=prices[0];
for(int i=1;i<prices.size();i++)
{
    if(prices[i]<minprofit)
    minprofit=prices[i];
    int money=prices[i]-minprofit;
    if(money>maxpro)
    maxpro=money;
}
return maxpro;
}
    };