Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer fee representing a transaction fee.
You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)
Return the maximum profit you can make.
Example 1:
Input: prices = [1, 3, 2, 8, 4, 9], fee = 2 Output: 8 Explanation: The maximum profit can be achieved by:
- Buying at prices[0] = 1
- Selling at prices[3] = 8
- Buying at prices[4] = 4
- Selling at prices[5] = 9
The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
Note:
0 < prices.length <= 50000.0 < prices[i] < 50000.0 <= fee < 50000.
解题思路：
在一段数组中，如果遇到价格更低的，那么买入价要变化，如果遇到价格高的，那么要拿利润与之前的比较是否更多，那么什么时候确定这段结束（卖出），重新买入下一个价格呢，就是当时的新的价格-最大利润，如果这个价格比之前的最小买入要低，那么重新开始。
class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
if(prices.size()==0||prices.size()==1) return 0;
int sum=0;
int min_buy=prices[0];
int max_profit=0;
for(int i=1;i<prices.size();i++){
if(prices[i]-min_buy-fee>max_profit) max_profit=prices[i]-min_buy-fee;
if(prices[i]<min_buy) min_buy=prices[i];
if(prices[i]-max_profit<=min_buy){sum+=max_profit;min_buy=prices[i];max_profit=0;}
}
sum+=max_profit;
return sum;
}
};