0122. Best Time to Buy and Sell Stock I I

# 122. Best Time to Buy and Sell Stock II#

## 题目 #

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 (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.
``````

## 解题思路 #

• 这一题是第 121 题的加强版。要求输出最大收益，这一题不止买卖一次，可以买卖多次，买卖不能在同一天内操作。
• 最大收益来源，必然是每次跌了就买入，涨到顶峰的时候就抛出。只要有涨峰就开始计算赚的钱，连续涨可以用两两相减累加来计算，两两相减累加，相当于涨到波峰的最大值减去谷底的值。这一点看通以后，题目非常简单。

## 代码 #

``````
package leetcode

func maxProfit122(prices []int) int {
profit := 0
for i := 0; i < len(prices)-1; i++ {
if prices[i+1] > prices[i] {
profit += prices[i+1] - prices[i]
}
}
return profit
}

``````

Apr 8, 2023