链接: 买卖股票的最佳时机

题目描述

算法分析

程序代码

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> f(n,vector<int>(3,-0x3f3f3f));
        auto g = f;
        f[0][0] = -prices[0];
        g[0][0] = 0;
        for(int i = 1;i < n;++i)
        {
            for(int j = 0;j < 3;++j)
            {
                f[i][j] = max(f[i-1][j],g[i-1][j]-prices[i]);
                g[i][j] = g[i-1][j];
                if(j-1 >= 0)
                {
                    g[i][j] = max(g[i][j],f[i-1][j-1]+prices[i]);
                }
            }
        }

        return max(g[n-1][0],max(g[n-1][1],g[n-1][2]));
    }
};