Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Example 1:
Input: [7, 1, 5, 3, 6, 4] Output: 5 max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
Input: [7, 6, 4, 3, 1] Output: 0 In this case, no transaction is done, i.e. max profit = 0.
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size()<2) return 0;
int maxProfit = 0;
int minPrice = prices[0];
for(int i=1; i<prices.size(); i++)
{
if(prices[i]>prices[i-1])
maxProfit = std::max(maxProfit, prices[i]-minPrice);
else
minPrice = std::min(minPrice, prices[i]);
}
return maxProfit;
}
};
本文介绍了一种寻找股票交易中最大利润的算法。该算法通过一次遍历价格数组,跟踪最低购买价格并计算可能的最大利润。示例展示了如何使用该算法解决实际问题。
795

被折叠的 条评论
为什么被折叠?



