123. 买卖股票的最佳时机III

题目链接

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;
    }
};

 

posted @ 2021-04-01 10:45  洗盏更酌  Views(32)  Comments(0Edit  收藏  举报