LeetCode #714 Best Time to Buy and Sell Stock with Transaction Fee
2017-10-28 23:41
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题目
Your are given an array of integersprices, for which the
i-th element is the price of a given stock on day
i; and a non-negative integer
feerepresenting 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.
解题思路
采用动态规划的方法,一开始比较容易想到的思路是设置一个收益数组pro,令
pro[i]表示第
i天的最大收益,然后考虑
pro[i]与
pro[i - 1, i - 2 ... 0]的关系。在第
i天,我们只有三个操作:
不买也不卖。此时
pro[i] = pro[i - 1]。
买。此时
pro[i] = pro[i - 1] - prices[i]。
卖。此时
pro[i] = pro[i - 1] + prices[i] - fee。
这样,似乎可以天真地认为一个动态转移方程就出来了:
pro[i]=max{pro[i], pro[i−1]−prices[i], pro[i−1]+prices[i]−fee}
然而这个方程根本不可行,因为 pro[i−1]>pro[i−1]−prices[i] 是肯定成立的,也就是说利用上面的动态转移方程将永远也执行不了买股票的操作!
上面的思路行不通,根本原因是“买股票”和“卖股票”这两个操作是无法单独地衡量其收益的,这一次的“卖”操作必须减去上一次的“买”操作才能衡量收益,同理,这一次的“买”操作也必须结合上一次的“卖”操作来衡量收益。因此,我们需要维护两种第
i天的最大收益:一种是第
i天执行“买”操作的最大收益
buy_pro[i],另一种是第
i天执行“卖”操作的最大收益
sell_pro[i],这样,动态转移方程就变成了:
buy_pro[i]=max{buy_pro[i−1], sell_pro[i−1]−prices[i]}sell_pro[i]=max{sell_pro[i−1], prices[i]+buy_pro[i−1]−fee}
这里要特别注意
buy_pro[i]表示的是第
i天执行“买”操作的最大收益,而不是花费,因此有 sell_pro[i]=max{sell_pro[i−1], prices[i]+buy_pro[i−1]−fee} 而不是 sell_pro[i]=max{sell_pro[i−1], prices[i]−buy_pro[i−1]−fee} 。计算到最后时,只需比较
sell_pro和
buy_pro的最后一个元素哪个较大,选择最大的收益返回即可。
上述的动态转移方程需要 O(n) 的空间复杂度,但是观察可以发现,第
i天的操作的收益只与第
i - 1天的操作收益有关,因此根本不需要维护这样的两个数组,只要使用两个变量
buy_pro和
sell_pro即可。
C++代码实现
class Solution { public: int maxProfit(vector<int>& prices, int fee) { // 初始化,第一天只能买,不能卖 int buy_profit = -prices[0], sell_profit = 0; for (int i = 1; i < prices.size(); ++i) { // 如果 buy_profit 被改变, // 即 buy_profit = sell_profit - prices[i], // 则 sell_profit 必然不会改变,反之亦然。 // buy_profit 和 sell_profit 之中只有一个能改变, // 保证了两者之中必有一个是前一天的最大收益值, // 这是不用维护两个数组,降低空间复杂度的关键! buy_profit = max(buy_profit, sell_profit - prices[i]); sell_profit = max(sell_profit, prices[i] + buy_profit - fee); } return max(buy_profit, sell_profit); } };
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