要找一个值，减去前面最小的，那么就得到最大利润了

    2. class Solution {
    3. public:
    4.     int maxProfit(vector<int>& prices) {
    5.         if(!prices.size()) return 0;
    6.         int minval = prices[0];
    7.         int res = 0;
    8.         for(int i = 1; i < prices.size(); i++)
    9.         {
    10.             int tmp = prices[i];
    11.             prices[i] = tmp - minval;
    12.             minval = min(minval, tmp);
    13.             res = max(res, prices[i]);
    14.         }
    15.         return res;
    16.     }
    17. };

    第二次写题

    1. class Solution {
    2. public:
    3.     int maxProfit(vector<int>& prices) {
    4.         if(!prices.size()) return 0;
    5.         vector<int> res(prices.size());
    6.         int minval = prices[0];
    7.         int maxval = INT_MIN;
    8.         for(int i = 0; i < prices.size(); i++)
    9.         {
    10.             res[i] = prices[i] - minval;
    11.             minval = min(minval, prices[i]);
    12.             maxval = max(maxval, res[i]);
    13.         }
    14.         return maxval;
    15.     }
    16. };