forked from JojoYang666/Leetcode-1-300
-
Notifications
You must be signed in to change notification settings - Fork 0
/
_122_BestTimetoBuyandSellStockII.java
41 lines (37 loc) · 1.55 KB
/
_122_BestTimetoBuyandSellStockII.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
package leetcode_1To300;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _122_BestTimetoBuyandSellStockII {
/**
* 122. Best Time to Buy and Sell Stock II
* 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 as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However,
you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
case : [5, 1, 2, 3, 4]
time : O(n);
space : O(1);
* @param prices
* @return
*/
public static int maxProfit(int[] prices) {
if (prices == null || prices.length < 2) {
return 0;
}
int profit = 0;
for (int i = 1; i < prices.length; i++) {
if (prices[i] > prices[i - 1]) {
profit += prices[i] - prices[i - 1];
}
}
return profit;
}
}