leetcode714 - Best Time to Buy and Sell Stock with Transaction Fee - medium

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.

 

法1：DP。O(n^2) - TLE
遍历，每个点都看一看这次不动作，或者最后一次买卖是之前第m天到今天，这几种情况哪种收益最好。
递推公式：dp[i] = Math.max(dp[i], dp[m] + prices[i - 1] - prices[m - 1] - fee); 对所有[1, i) 的m循环

 

法2：DP。O(n)
通过转化memo记录内容的形式，让卖股票的时候不需要care这只股票是哪一天买入的，从而避免了前面的m循环，去遍历前面是哪一天买入的导致的O(n^2)。
定义：hold[i], sold[i] 表示第I天手上有股票和没股票分别能最多揣着多少钱。（注意不是profit的感觉，而是手头现金的感觉）最后返回sold[last]
初始化：hold[0] = -prices[0]; 注意第一天如果拿着股票的话你要先垫付。
递推公式：hold[i] = max(hold[i - 1], sold[i - 1] - prices[i]); sold[i] = Math.max(sold[i - 1], hold[i - 1] + prices[i] - fee);
解释：今天的情况由昨天的决定。比如今天手上有股票，可以由昨天有股票转化而来，或者由昨天没股票转化而来。今天手上没股票同理。

 

法1实现 O(n^2) TLE：

class Solution {
    public int maxProfit(int[] prices, int fee) {
        if (prices == null || prices.length == 0) {
            return 0;
        }
        int[] dp = new int[prices.length + 1];
        for (int i = 2; i < dp.length; i++) {
            for (int m = 1; m < i; m++) {
                dp[i] = Math.max(dp[i], dp[m] + prices[i - 1] - prices[m - 1] - fee);
            }
        }
        return dp[dp.length - 1];
    }
    
}

 

法2实现 O(n):

class Solution {
    public int maxProfit(int[] prices, int fee) {
        if (prices == null || prices.length == 0) {
            return 0;
        }
        
        int[] hold = new int[prices.length];
        int[] sold = new int[prices.length];
        hold[0] = -prices[0];
        
        for (int i = 1; i < prices.length; i++) {
            hold[i] = Math.max(hold[i - 1], sold[i - 1] - prices[i]);
            sold[i] = Math.max(sold[i - 1], hold[i - 1] + prices[i] - fee);
        }
        return sold[sold.length - 1];
    }
}