LeetCode Best Time to Buy and Sell Stock-优快云博客

本文探讨了在给定股票价格序列的情况下，如何通过一次买卖操作获取最大利润的算法。通过局部最优和全局最优的概念，实现了一个时间复杂度为O(n)且空间复杂度为O(1)的解决方案。

原题链接在这里：https://leetcode.com/problems/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.

题解：

与Maximum Subarray相似，都是全局最优与局部最优问题。

找到一段subarray, 找出最大profit. 如果profit已经小于0了, reset as 0.

local = Math.max(local + diff, 0). 0表示i天股价没有i-1天高. 只可以进行一次交易.

global = Math.max(local, global).

Time Complexity: O(n). Space: O(1).

AC Java:

 1 public class Solution {
 2     public int maxProfit(int[] prices) {
 3         //采用局部最优和全局最优
 4         if(prices == null || prices.length <= 1){
 5             return 0;
 6         }
 7         int local = 0;
 8         int global = 0;
 9         for(int i = 1; i<prices.length; i++){
10             local = Math.max(local+prices[i]-prices[i-1],0);
11             global = Math.max(local,global);
12         }
13         return global;
14     }
15 }