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).
网上有解法是把所有的升值都加起来,我觉得不好解释,其实那当然是最优的了。
这个实现其实也可以用local 最低和最高来算。。。
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n=prices.size();
if (n<=1)
return 0;
int total=0;
for (int i=0; i<n; i++){
int myMin=0;
int myMax=0;
for (; i<n-1 && prices[i]>prices[i+1]; i++);
myMin=prices[i];
for (; i<n-1 && prices[i]<prices[i+1]; i++);
myMax=prices[i];
total+= myMax-myMin;
}
return total;
}
};
本文介绍了一种在股票市场中通过多次买卖实现最大利润的算法。该算法利用局部最低和最高价格计算利润,允许无限次交易但禁止同时持有股票。
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