leetcode 122.买卖股票的最佳时机Ⅱ

　　贪心。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int now=prices.front();  //现在手中买入价
        int profit=0;
        prices.push_back(prices.back());
        for (int i=0;i<prices.size()-1;i++)
            if (prices[i+1]<prices[i])    //明天将跌
                profit+=(prices[i]-now),now=prices[i+1];
        return profit+prices.back()-now;
    }
};

标签：vector,int,back,maxProfit,profit,122,最佳时机,prices,leetcode
来源： https://www.cnblogs.com/wegret/p/14872450.html