LeetCode Best Time to Buy and Sell Stock III

最新推荐文章于 2022-07-03 21:43:22 发布

转载最新推荐文章于 2022-07-03 21:43:22 发布 · 36 阅读

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原文链接：http://www.cnblogs.com/birdhack/p/4156599.html

本文探讨了一个在给定股票价格数组中寻找最大利润的算法，限制最多进行两次交易，且不能同时进行买卖操作。通过创建两个辅助数组分别记录从左到右和从右到左的最大利润，最终通过遍历找到最大总利润。

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Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

 1 public class Solution {
 2     public int maxProfit(int[] prices) {
 3         if (prices.length<=1) {
 4             return 0;
 5         }
 6 
 7         int[] left = new int[prices.length];
 8         int[] right = new int[prices.length];
 9         int min = prices[0];
10         for (int i = 1; i < prices.length; i++) {
11             left[i] = left[i - 1] > prices[i] - min ? left[i - 1] : prices[i] - min;
12             min = min < prices[i] ? min : prices[i];
13         }
14 
15         int max = prices[prices.length - 1];
16         for (int i = prices.length - 2; i >= 0; i--) {
17             right[i] = right[i + 1] > max - prices[i] ? right[i + 1] : max - prices[i];
18             max = max > prices[i] ? max : prices[i];
19         }
20         int sum = Integer.MIN_VALUE;
21         for (int i = 0; i < left.length; i++) {
22             sum = sum > left[i] + right[i] ? sum : left[i] + right[i];
23         }
24         return sum;
25     }
26 }