LeetCode: Best Time to Buy and Sell Stock

思路：记录到当前位置i为止，最小值的位置 min_index， 则到当前位置最大收益就是max( 已有收益， prices[i] - prices[index])。

code:

class Solution {public:    int maxProfit(vector<int> &prices) {        int len = prices.size();        if(len == 0)return 0;        int min_v = prices[0], min_index = 0;        int ret = 0;        for(int i = 1;i < len;i++){            if(min_v > prices[i]){                min_v = prices[i];                min_index = i;            }            ret = max(ret,prices[i] - prices[min_index]);        }        return ret;    }};