122. Best Time to Buy and Sell Stock II

题目

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).

分析

由于可以进行多次交易，故顺次遍历价格向量，遵循低买高卖的原则即可。

class Solution {public:    int maxProfit(vector<int>& prices) {        if(prices.empty())            return 0;        int buy=-1;        int maxProfit=0;        for(int i=0;i<prices.size()-1;++i)        {            if(prices[i]<prices[i+1]&&buy==-1)//当前价格低于其后价格时买入                buy=prices[i];            if(prices[i]>prices[i+1]&&buy!=-1)//当前价格高于其后价格时卖出            {                maxProfit+=prices[i]-buy;                buy=-1;            }        }        if(buy!=-1&&prices[prices.size()-1]>buy)//考虑最后一个元素是否卖出            maxProfit+=prices[prices.size()-1]-buy;        return maxProfit;    }};