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) {
int cost = 1000000,ans = 0;
for(int x : prices){
ans = max(ans,x-cost);
if(cost > x)
cost = x;
}
return ans;
}
};
本文介绍了一种寻找股票交易中最大利润的算法。该算法通过遍历价格数组,跟踪最低买入价格并计算当前价格与最低买入价格之间的差值来确定最大利润。适用于只允许进行一次买卖的情况。
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