123. 买卖股票的最佳时机III
解题思路:动态规划
C++:
class Solution { public: int maxProfit(vector<int>& prices) { int n = prices.size(); int buy1 = -prices[0], sell1 = 0; int buy2 = -prices[0], sell2 = 0; for (int i = 1; i < n; ++i) { buy1 = max(buy1, -prices[i]); sell1 = max(sell1, buy1 + prices[i]); buy2 = max(buy2, sell1 - prices[i]); sell2 = max(sell2, buy2 + prices[i]); } return sell2; } };