letcode 714. Best Time to Buy and Sell Stock with Transaction Fee
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.
給定一個數組,陣列中的第i個元素為一隻股票第i天的價格。現在求買賣股票所能得到的最大值。 如果用貪心的思想去解這道題,應注意 貪心選擇 與 產生的子問題無關。因此,只能確定應該在何時買入,而無法確定何時賣出(因為沒有標準去衡量賺多少錢是最優的)。所以,我們需要制定一個判斷標準去衡量應該在何時買入。 如果對於一個單調上升的數列,其最優解必然是 最大值 - 最小值 。但如果指定的區間內不是單調上升的,那麼假定存在一個 比區間最小值大的某個值,和 比區間最大值小的某個值 ,能將這個區間分成兩部分,並且使這兩部分的各自的 最大值 - 最小值 的和 大於 原區間 最大值 - 最小值。
class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int len=prices.size();
if(len==0 || len==1)
return 0;
int sum=0;
int peak=0;
int least=INT_MAX;
for(auto ele : prices){
if(peak-ele>fee){
sum+=(peak-least-fee);
least=ele;
peak=0;
}
least=min(ele,least);
if(ele-least>fee){
peak=max(peak,ele);
}
}
if(peak-least>fee)
sum+=(peak-least-fee);
return sum;
}
};
注意:
- 由於是判斷何時買入,所以在執行買入前先執行賣出操作
- 賣出操作時注意對下個區間最大值和最小值的初始化
- 由於判斷何時買入,在沒有新的買入前不進行賣出,因此在迴圈結束時,要進行最後一次賣出