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.

思路：一次扫描即可。只需要找到最大增长即可。max(prices[j] – prices[i]) ，i < j

从前往后，用当前价格减去此前的最低价格，就是在当前点卖出股票能获得的最高利润。

扫描中，更新最高利润和最低价格即可。

复杂度：O(N)

AC Code:

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        //遍历，目标找到一对依次出现的最低点和最高点，使得满足达到最大利润
        
        if(prices.size() <= 1)
            return 0;
        
        int low = prices[0];
        int maxprofit = 0;
        
        for(int i = 1; i < prices.size(); i++)
        {
            int profit = prices[i] - low;
            if(maxprofit < profit) maxprofit = profit;
            if(prices[i] < low) low = prices[i];
        }       
        return maxprofit;
    }
};

[C++]LeetCode: 54 Best Time to Buy and Sell Stock