714. Best Time to Buy and Sell Stock with Transaction Fee

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] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

Note:

• 0 < prices.length <= 50000.
• 0 < prices[i] < 50000.
• 0 <= fee < 50000.
• 
  

  分析

  第i天的最大现金，等于第i-1天的没有买股票时的最大现金、和第i-1的买了股票的最大现金+股价-交易费，中的最大者。

  第i天买了股票前提下的最大现金，等于第i-1天的买了股票的最大现金、和第i-1天的没有买股票的最大现金-股票，中的最大者

  
  

  code

  #include <iostream>
  #include <string>
  #include <vector>
  #include <algorithm>
  #include <numeric>
  using namespace std;
  class Solution {
  public:
      int maxProfit(vector<int>& prices, int fee) {
      	int n=prices.size();
          int sell[n]={},buy[n]={};
          sell[0]=0,buy[0]=-prices[0];
          for (int i = 1; i < n; ++i) {
          	sell[i]=max(sell[i-1],buy[i-1]+prices[i]-fee);
          	buy[i]=max(buy[i-1],sell[i-1]-prices[i]);
  		}
  		return sell[n-1];
      }
  };
  int main(int argc, char const *argv[])
  {
  	int a[]={1, 3, 2, 8, 4, 9};
  	vector<int> nums(a,a+6);
  	Solution s;
  	cout << s.maxProfit(nums,2);
  	return 0;
  }