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

时间:2017-11-25 00:34:20      阅读:63      评论:0      收藏:0      [点我收藏+]

标签:pla   tor   before   which   each   code   pri   let   xpl   

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.

 

用maxProfit[i]记录前i个数据的最大收益。

用cashWhenHoldOne[i]表示对于前i个数据,在手中有且只有一个stock的情况下,手中现金的最大值。

则:

maxProfit[i] = max(maxProfit[i-1], cashWhenHoldOne[i-1] + prices[i] - fee);
cashWhenHoldOne[i] = max(cashWhenHoldOne[i-1], maxProfit[i-1] - prices[i]);

 

我的代码如下:

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int len = prices.size();
        vector<int> maxProfit(len), cashWhenHoldOne(len);
        maxProfit[0] = 0, cashWhenHoldOne[0] = -prices[0];
        for (int i = 1; i < len; i++) {
            maxProfit[i] = max(maxProfit[i-1], cashWhenHoldOne[i-1] + prices[i] - fee);
            cashWhenHoldOne[i] = max(cashWhenHoldOne[i-1], maxProfit[i-1] - prices[i]);
        }
        return maxProfit[len-1];
    }
};

 

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

标签:pla   tor   before   which   each   code   pri   let   xpl   

原文:http://www.cnblogs.com/zmj97/p/7892727.html

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