121. Best Time to Buy and Sell Stock 类别：动态规划难度：easy

原创于 2016-12-26 16:09:40 发布 · 153 阅读

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本文介绍了一种寻找股票交易中最大利润的算法。通过遍历价格数组，算法记录最低购买价格并在每次迭代中更新最大利润，确保卖出价格高于买入价格。示例展示了如何计算最大可能利润。

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题目：

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.

思路：

记ret为收益，随着prices的遍历，如果当前遇到的元素比curMin小，就重新更新curMin，计算prices[i]与curMin的差值，并更新ret。

程序：

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