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  • 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;

    相关题

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    买卖股票的最佳时间2 LeetCode122 https://www.cnblogs.com/zhacai/p/10596627.html

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  • 原文地址:https://www.cnblogs.com/zhacai/p/10659288.html
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