LeetCode-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 feerepresenting 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.

使用dp 时间复杂度为O(n) ，只记录上一个点的信息，即空间复杂度为O(1)

 1  public int maxProfit(int[] prices, int fee) {
 2         if(null==prices||0==prices.length){
 3             return 0;
 4         }
 5         int stock =-prices[0]-fee;
 6         int noStock =0;
 7         int preNoStock=0;
 8         for(int i=1;i<prices.length;i++){
 9             preNoStock=noStock;
10             noStock=Math.max(noStock,stock+prices[i]);//第i天无股票是第i-1天无股票和第i天卖股票的最大值
11             stock=Math.max(preNoStock-prices[i]-fee,stock);//第i天有股票是第i-1天有股票和第i天买股票的最大值
12         }
13         return noStock;