Update 122.best-time-to-buy-and-sell-stock-ii.md

Author: azl397985856

problems/122.best-time-to-buy-and-sell-stock-ii.md

@@ -54,62 +54,6 @@ Explanation: In this case, no transaction is done, i.e. max profit = 0.
 JS Code:
 
 ```js
-
-/*
- * @lc app=leetcode id=122 lang=javascript
- *
- * [122] Best Time to Buy and Sell Stock II
- *
- * https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/
- *
- * algorithms
- * Easy (50.99%)
- * Total Accepted:    315.5K
- * Total Submissions: 610.9K
- * Testcase Example:  '[7,1,5,3,6,4]'
- *
- * Say you have an array for which the i^th 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 (i.e., buy one and sell one share of the stock
- * multiple times).
- * 
- * Note: You may not engage in multiple transactions at the same time (i.e.,
- * you must sell the stock before you buy again).
- * 
- * Example 1:
- * 
- * 
- * Input: [7,1,5,3,6,4]
- * Output: 7
- * Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit
- * = 5-1 = 4.
- * Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 =
- * 3.
- * 
- * 
- * Example 2:
- * 
- * 
- * Input: [1,2,3,4,5]
- * Output: 4
- * Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit
- * = 5-1 = 4.
- * Note that you cannot buy on day 1, buy on day 2 and sell them later, as you
- * are
- * engaging multiple transactions at the same time. You must sell before buying
- * again.
- * 
- * 
- * Example 3:
- * 
- * 
- * Input: [7,6,4,3,1]
- * Output: 0
- * Explanation: In this case, no transaction is done, i.e. max profit = 0.
- * 
- */
 /**
  * @param {number[]} prices
  * @return {number}
@@ -137,7 +81,6 @@ class Solution:
         gains = [prices[i] - prices[i-1] for i in range(1, len(prices))
                  if prices[i] - prices[i-1] > 0]
         return sum(gains)
-print(Solution().maxProfit([7, 1, 5, 3, 6, 4]))
 #评论区里都讲这是一道开玩笑的送分题.
 ```