Description
Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most two transactions.
Solution
动态规划法：用两个数组，数组f1[i]表示在[0, i]范围内进行一次买入卖出的最大收益，数组f2[i]表示在[i, n-1]范围内进行一次买入卖出的最大收益，则总的最大收益为max(f1[i]+f2[i])，数组f1、f2的求法参考题目121
python
1 class Solution(object): 2 def maxProfit(self, prices): 3 """ 4 :type prices: List[int] 5 :rtype: int 6 """ 7 length = len(prices) 8 if length <= 1: 9 return 0 10 11 f1 = [] 12 f2 = [] 13 min_price = prices[0] 14 f1.append(0) 15 for i in range(1, length): 16 min_price = min(min_price, prices[i]) 17 f1.append(max(f1[i-1], prices[i]-min_price)) 18 19 # reverse the prices list to iterate them from behind 20 prices_re = prices 21 prices_re.reverse() 22 f2.append(0) 23 max_price = prices_re[0] 24 for i in range(1, length): 25 max_price = max(max_price, prices_re[i]) 26 f2.append(max(f2[i-1], max_price-prices_re[i])) 27 28 f2.reverse() 29 max_profit = 0 30 for i in range(length): 31 max_profit = max(max_profit, f1[i]+f2[i]) 32 33 return max_profit
cpp
1 class Solution { 2 public: 3 int maxProfit(vector<int>& prices) { 4 int length = prices.size(); 5 if (length <= 1) 6 return 0; 7 8 vector<int> f1(length); 9 vector<int> f2(length); 10 f1[0] = 0; 11 int min_price = prices[0]; 12 // get max_profit1 using positive sequence 13 for (int i=1; i<length; ++i) 14 { 15 min_price = min(min_price, prices[i]); 16 f1[i] = max(f1[i-1], prices[i]-min_price); 17 } 18 19 // get max_profit2 using inverted sequence 20 f2[length-1] = 0; 21 int max_price = prices[length-1]; 22 for (int i=length-2; i>=0; --i) 23 { 24 max_price = max(max_price, prices[i]); 25 f2[i] = max(f2[i+1], max_price-prices[i]); 26 } 27 28 // generate max_profit for output 29 int max_profit = 0; 30 for (int i=0; i<length; ++i) 31 { 32 max_profit = max(max_profit, f1[i]+f2[i]); 33 } 34 35 return max_profit; 36 } 37 };