Leetcode 121. 买卖股票的最佳时机

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.size()==0)
            return 0;
        int maxProfit = 0x80000000;
        vector<int> stack;
        stack.push_back(prices[0]);
        for(int i=1;i<prices.size(); ++i)
        {
            if(stack.back() < prices[i])
            {
                stack.push_back(prices[i]);
            }
            else
            {
                maxProfit = max(maxProfit, stack.back()-stack.front());
                stack.pop_back();
                while(stack.size()>0)
                {
                    auto tmp = stack.back();
                    if(tmp >= prices[i])
                    {
                        stack.pop_back();
                    }
                    else
                    {
                        break;
                    }
                }
                stack.push_back(prices[i]);
            }
        }
        if(stack.size() == 1)
        {
            maxProfit = max(maxProfit, 0);
        }
        else
        {
            maxProfit = max(maxProfit, stack.back()-stack.front());
        }
        return maxProfit;
    }
};

原文地址：https://www.cnblogs.com/randyniu/p/9374752.html