leetcode:(121) Best Time to Buy and Sell Stock(java)

最新推荐文章于 2022-03-15 16:07:17 发布

Sunshine_liang1

最新推荐文章于 2022-03-15 16:07:17 发布

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本文链接：https://blog.youkuaiyun.com/Sunshine_liang1/article/details/83993166

本文介绍了一种通过记录最低买入价并计算与后续卖出价之间的差额，来寻找股票交易中最大利润的算法。该算法适用于只允许进行一次交易的情况，通过遍历股票价格数组，动态更新最低买入价和最大利润。

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题目描述：

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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 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
Explanation: In this case, no transaction is done, i.e. max profit = 0.

解题思路：

只要记录前面的最小价格，将这个最小价格作为买入价格，然后将当前的价格作为售出价格，查看当前收益是不是最大收益。

具体思路及代码如下：

package Leetcode_Github;

public class GreedyThought_MaxProfit_121_1112 {
    public int MaxProfit(int[] prices){
        if (prices == null || prices.length == 0) {
            return 0;
        }
        int minPrice = prices[0];
        int maxProfit = 0;
        for (int i = 1; i < prices.length; i++) {
            if (prices[i] < minPrice) {
                minPrice = prices[i];
            } else {
                maxProfit = Math.max(maxProfit, prices[i] - minPrice);
            }
        }
        return maxProfit;
    }

    public static void main(String[] args) {
        int[] nums = {7, 1, 5, 3, 6, 4};
        GreedyThought_MaxProfit_121_1112 test = new GreedyThought_MaxProfit_121_1112();
        int result = test.MaxProfit(nums);
        System.out.println(result);
    }
}