Best Time to Buy and Sell Stock最大收益问题算法详解

问题详见： Best Time to Buy and Sell Stock

题目让我们求解一个给定的数组中购买和出售（必须在购买之后）所能获得的最大收益。题目描述如下：
      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.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5
max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]
Output: 0
In this case, no transaction is done, i.e. max profit = 0.

解题思路：

      由题可知，只要找到位置靠后的数组元素与靠前的一个数组元素的差值最大的差值就是所求最大收益。所以先判断是否为空，为空肯定是0，然后看遍历到的元素与已经遍历到的元素中的最小值的大小和差值，动态更新遍历到的最小元素和最大收益。整个算法复杂度为O(n)。具体算法如下：

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

其提交运行结果如下：