【Leetcode-Easy-121】Best Time to Buy and Sell Stock

最新推荐文章于 2020-12-17 23:26:32 发布

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最新推荐文章于 2020-12-17 23:26:32 发布

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本文介绍了一种寻找股票买卖最佳时机以获得最大利润的算法。该算法通过跟踪历史最低价和当前价格之间的最大差值来确定最佳买卖点。示例展示了如何在不同市场条件下应用此算法。

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【Leetcode-Easy-121】Best Time to Buy and Sell Stock

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

找出截止当前位置的最小值和截至当前位置的最大值，
记录两者之差，保留最大的差值。

程序

class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null || prices.length == 0) return 0;
        int profit = 0;
        int low = prices[0];
        for (int i = 0; i < prices.length; i ++){
            low = Math.min(low, prices[i]);
            profit = Math.max(profit, prices[i]-low);
        }
        return profit;
    }
}

声明：题目来自Leetcode.