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 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
.
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;
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