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

題目

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.

思路

第i天有兩種情況：持有股票及沒有股票。對於持有股票，有兩種操作：賣，或者不賣；對於沒有股票，也有兩種操作：買，或者不買。用sell[i]表示第i天沒有股票（或者執行“賣”操作）所能獲得的最大利潤，buy[i]表示第i天持有股票（或者執行“買”操作）所能獲得的最大利潤，則有：

sell[i] = max(sell[i-1], buy[i-1]+prices[i]-fee)
buy[i] = max(buy[i-1], sell[i-1]-prices[i])

程式碼

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

後

太久沒打演算法題啥都不會，，，想了半天最後還是看了網上解法才略懂了，sad