LeetCode - Best Time to Buy and Sell Stock

Question

Link : https://leetcode.com/problems/best-time-to-buy-and-sell-stock/

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

说明:
哈哈哈哈，可能英语不好刚开始看到题目时候不是很清楚题目的意思。后来发现题意是这样的：有一个数组，分别存了若干个整数代表股票的价格，现在需要你找出两个时刻分别代表股票买入和卖出的价格，并且使得收益最大。其中一个隐含的要求自然就是要先买入才能卖出了哈哈哈哈。

Code

一个比较挫的版本，时间复杂度为O(n2)，并且会超时不通过的。不过这是最简单的实现版本，就放上来了。

class Solution {public:    int maxProfit(vector<int>& prices) {        if(prices.size() == 0)            return 0;        int profit = 0;        for(int i = 0; i < prices.size(); i++){            for(int j = i + 1; j < prices.size(); j++){                int pro = prices[j] - prices[i];                if(pro > profit)                    profit = pro;            }        }        return profit;    }};

下面这是时间复杂度为O(n)的版本，并且是可以通过的。(C++ : 8ms)

class Solution {public:    int maxProfit(vector<int>& prices) {        if(prices.size() == 0)            return 0;        int profit = 0, brought = prices[0];        for(int i = 1; i < prices.size(); i++){            if(brought < prices[i]){                int pro = prices[i] - brought;                if(pro > profit)                    profit = pro;            }            else                brought = prices[i];        }        return profit;    }};