【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).
【题意】
给定一个数组prices, prices[i]表示第i天的股价。本题没有买卖次数的限制，问最大收益是多少
【思路】
贪心思想。假设股价一路上扬，则持币观望，直到股价升到最高点抛售获利最大。
【代码】
class Solution {
public:
int maxProfit(vector<int> &prices) {
int size=prices.size();
if(size<=1)return 0;
int maxProfit=0;
int buyDate=0;
for(int i=1; i<size; i++){
if(prices[i-1]>prices[i]){
//股价開始下跌，在i-1天抛售
maxProfit+=prices[i-1]-prices[buyDate];
buyDate=i;
}
}
//【注意】别忘了最后一次买卖
maxProfit+=prices[size-1]-prices[buyDate];
return maxProfit;
}
};