diff options
Diffstat (limited to 'buy_sell_stock_fee.cpp')
-rw-r--r-- | buy_sell_stock_fee.cpp | 35 |
1 files changed, 35 insertions, 0 deletions
diff --git a/buy_sell_stock_fee.cpp b/buy_sell_stock_fee.cpp new file mode 100644 index 0000000..0375c56 --- /dev/null +++ b/buy_sell_stock_fee.cpp @@ -0,0 +1,35 @@ +// 714. Best Time to Buy and Sell Stock with Transaction Fee +#include "leetcode.h" + +/* + * given an array 'prices' where 'prices[i]' is the price of a given stock on + * the i'th day, and an integer 'fee' representing a transaction fee. find the + * maximum profit you can achieve. you may complete as many transactions as youl + * like, but you need to pay the transaction fee for each transaction. note: you + * may not engage in multiple transactions simultaneously (ie. you must sell the + * stock before you buy again) + */ + +class Solution { +public: + int maxProfit(vector<int> &prices, int fee) { + int n = prices.size(); + if (n < 2) + return 0; + vector<int> hold(n, 0), sold(n, 0); + hold[0] = -prices[0]; + for (int i = 1; i < n; i++) { + hold[i] = max(hold[i - 1], sold[i - 1] - prices[i]); + sold[i] = max(sold[i - 1], hold[i - 1] - fee + prices[i]); + } + return sold[n - 1]; + } +}; + +int main() { + Solution obj; + vector<int> p1 = {1, 3, 2, 8, 4, 9}; + vector<int> p2 = {1, 3, 7, 5, 10, 3}; + printf("%d\n", obj.maxProfit(p1, 2)); // expect: 8 + printf("%d\n", obj.maxProfit(p2, 3)); // expect: 6 +} |