class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int result = 0;
int minPrice = prices[0]; // 记录最低价格
for (int i = 1; i < prices.size(); i++) {
// 情况二:相当于买入
if (prices[i] < minPrice) minPrice = prices[i];
// 情况三:保持原有状态(因为此时买则不便宜,卖则亏本)
if (prices[i] >= minPrice && prices[i] <= minPrice + fee) {
continue;
}
// 计算利润,可能有多次计算利润,最后一次计算利润才是真正意义的卖出
if (prices[i] > minPrice + fee) {
result += prices[i] - minPrice - fee;
minPrice = prices[i] - fee; // 情况一,这一步很关键
}
}
return result;
}
};
Time Complexity: O(n)
Space Complexity: O(1)