题目链接:121. 买卖股票的最佳时机 - 力扣(LeetCode)
就是要今日的股价和历史股价的最大差
class Solution {
public:
int maxProfit(vector<int> &prices) {
int min = prices[0], profit = 0;
for (int i = 1; i < prices.size(); i++) {
min = std::min(prices[i], min);
profit = max(profit, prices[i] - min);
}
return profit;
}
};