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.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
设置一个指针，分别求出前面的最大收益与后面的最大收益，将两者相加，可以提前计算好存在两个数组里以避免重复计算。
1 class Solution { 2 public: 3 int maxProfit(vector<int> &prices) { 4 int n = prices.size(); 5 if (n == 0) return 0; 6 vector<int> first(n, 0); 7 vector<int> second(n, 0); 8 int max_profit = 0, cur_min = prices[0], cur_max = prices[n-1]; 9 for (int i = 1; i < n; ++i) { 10 max_profit = max(max_profit, prices[i] - cur_min); 11 cur_min = min(cur_min, prices[i]); 12 first[i] = max_profit; 13 } 14 max_profit = 0; 15 for (int i = n - 2; i >= 0; --i) { 16 max_profit = max(max_profit, cur_max - prices[i]); 17 cur_max = max(cur_max, prices[i]); 18 second[i] = max_profit; 19 } 20 max_profit = 0; 21 for (int i = 0; i < n; ++i) { 22 max_profit = max(max_profit, first[i] + second[i]); 23 } 24 return max_profit; 25 } 26 };
下面是Best Time to Buy and Sell Stock I的代码，有人说将问题转化为最大连续子序列和的问题，其实不用那样，只要保存当前最小值就可以解决问题。
1 class Solution { 2 public: 3 int maxProfit(vector<int> &prices) { 4 int n = prices.size(); 5 if (n == 0) return 0; 6 int max_profit = 0, cur_min = prices[0]; 7 for (int i = 0; i < n; ++i) { 8 max_profit = max(max_profit, prices[i] - cur_min); 9 cur_min = min(cur_min, prices[i]); 10 } 11 return max_profit; 12 } 13 };