2018年6月1日 星期五

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

轉自LeetCode

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.


  • <Solution>


    想法如下
    • 用 cash 表示買入股票後,手上的現金
    • 用 profit 表示目前手上的現金,也就是最後的獲利
    • 歷遍 prices,計算今天股票要不要交易,cash 和 profit 的變化,取最大值
    code 如下


    Kotlin

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