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).
Analysis:
A bit different with previous one. Since we can make unlimited transactions, this question turns to sum all the positive
price difference.
So, scan from left to right, and add all positive diff value.
Java
public int maxProfit(int[] prices) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
int maxPro = 0;
for(int i =1; i<prices.length;i++){
int delta = prices[i] - prices[i-1];
if(delta > 0)
maxPro += delta;
}
return maxPro;
}c++
int maxProfit(vector<int> &prices) {
int sum = 0;
for(int i=1;i<prices.size();i++){
if(prices[i]-prices[i-1] > 0)
sum+= prices[i]-prices[i-1];
}
return sum;
}
股票买卖算法

本文介绍了一种用于计算股票交易最大利润的算法。该算法允许无限次买卖操作,但必须在再次购买前卖出。通过扫描价格数组并累加所有正的价格差值来实现。
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