LeetCode 121. Best Time to Buy and Sell Stock

Das-Boot

于 2017-04-30 22:00:57 发布

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分类专栏： Algorithms 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/u010787069/article/details/71036705

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本文介绍了一种通过遍历股票价格数组寻找最大交易利润的方法。该算法使用两个变量记录遍历过程中的最小购买价格和最大利润，最终返回最大可能利润。适用于只允许进行一次买卖的情况。

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Description

Say you have an array for which the $i^{th}$ element is the price of a given stock on day .
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.

Solution

遍历一次数组，用一个变量记录遍历过数中的最小值，然后每次计算当前值和这个最小值之间的差值记为利润，每次选较大的利润来更新。当遍历完成后当前利润即为所求。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res = 0, buy = INT_MAX;
        for (int price : prices) {
            buy = min(buy, price);
            res = max(res, price - buy);
        }
        return res;
    }
};