[LeetCode] Best Time to Buy and Sell Stock with Transaction Fee 买股票的最佳时间含交易费

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[LeetCode] Best Time to Buy and Sell Stock with Transaction Fee 买股票的最佳时间含交易费

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.

sold[i] = max(sold[i - 1], hold[i - 1] + prices[i] - fee);

hold[i] = max(hold[i - 1], sold[i - 1] - prices[i]);

```public:
int maxProfit(vector<int>& prices, int fee) {
vector<int> sold(prices.size(), 0), hold = sold;
hold[0] = -prices[0];
for (int i = 1; i < prices.size(); ++i) {
sold[i] = max(sold[i - 1], hold[i - 1] + prices[i] - fee);
hold[i] = max(hold[i - 1], sold[i - 1] - prices[i]);
}
return sold.back();
}
}; ```

class Solution {
```public:
int maxProfit(vector<int>& prices, int fee) {
int sold = 0, hold = -prices[0];
for (int price : prices) {
int t = sold;
sold = max(sold, hold + price - fee);
hold = max(hold, t - price);
}
return sold;
}
};```

https://discuss.leetcode.com/topic/107992/java-dp-solution-easy-understand

https://discuss.leetcode.com/topic/107977/c-concise-solution-o-n-time-o-1-space

https://discuss.leetcode.com/topic/107998/most-consistent-ways-of-dealing-with-the-series-of-stock-problems

，如需转载请自行联系原博主。

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