Leetcode——121. Best Time to Buy and Sell Stock

题目原址

https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/

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 (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)

Example2:

Input: [7, 6, 4, 3, 1]
Output: 0
In this case, no transaction is done, i.e. max profit = 0.

解题思路

翻译：
你有一个数组，数组中的数是当天的股票价格，如prices[1] = 10，表示第2天的股票价格为10
你最多有一次交易的机会（即：只能买一次股票，卖一次股票）设计一个算法找出交易的最大利润。

这道题使用Kadane’s algorithm(Kadane算法)实现，Kadane是卡内基梅隆大学的教授，而该算法是为了解决最大子序列的和（maximum subarray）提出的。
Leetcode上面有很多求最大子序列和的问题，这种题应该算是一个类型的。
解决该题的方法还是很简单的：
- 判断当当前最大利益 + 当前的元素-当前元素的前一个元素的值是否是负数，如果是负数，就说明已经赔没了，不能买当前股票的前面的股票了，所以当前的利益为0。如果值为正数，则还可以往下遍历。
- 每次更改当前利益之后，都要判断当前利益与总利益之间的大小，保证总利益永远最大。

AC代码

class Solution {public int maxProfit(int[] prices) {int max_ending_here = 0;int max_so_far = 0;for(int i = 1; i < prices.length; i++) {max_ending_here = Math.max(0, prices[i] - prices[i-1] + max_ending_here);max_so_far = Math.max(max_so_far, max_ending_here);}return max_so_far;   }
}

感谢

https://discuss.leetcode.com/topic/19853/kadane-s-algorithm-since-no-one-has-mentioned-about-this-so-far-in-case-if-interviewer-twists-the-input/3