[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>

Best Time to Buy and Sell Stock 的衍生題

想法如下

• 可以沿用 122. Best Time to Buy and Sell Stock II 的想法，但要把 fee 考慮進來

• 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

Java(參考解答)

	class Solution {
	public int maxProfit(int[] prices, int fee) {
	int cash = Integer.MIN_VALUE; //>> the money you have after buying a stock
	int profit = 0; //>> the money you have after selling a stock
	for(int p : prices) {
	cash = Math.max(cash, profit - p); //>> max(no buy, buy today's stock);
	profit = Math.max(profit, cash + p - fee); //>> max(no sell, sell stock with today's price plus fee)
	}
	return profit;
	}
	}

view raw maxProfitWithFee.java hosted with ❤ by GitHub

Kotlin

	class Solution {
	fun maxProfit(prices: IntArray, fee: Int): Int {
	var cash = Int.MIN_VALUE // the money you have after buying a stock
	var profit = 0
	for(p in prices) {
	cash = Math.max(cash, profit - p) //max(do nothing, buy today)
	profit = Math.max(profit, cash + p - fee) //max(do nothing, sell today)
	}
	return profit
	}
	}

view raw best_time_to_buy_and_sell_stock_with_transaction_fee.kt hosted with ❤ by GitHub