# Best Time to Buy and Sell Stock II

## Question

> 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).
>
> Example
>
> Given an example \[2,1,2,0,1], return 2

**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.
```

## Analysis

很基本的转化问题，只需考虑所有相邻两天的增量是否为正值，如果为正值，则可以计入收益profit。

## Solution

```java
class Solution {
    /**
     * @param prices: Given an integer array
     * @return: Maximum profit
     */
    public int maxProfit(int[] prices) {
        if (prices == null || prices.length == 0) {
            return 0;
        }
        int profit = 0;
        for (int i = 0; i < prices.length - 1; i++) {
            int diff = prices[i + 1] - prices[i];
            if (diff > 0) {
                profit += diff;
            }
        }
        return profit;
    }
}
```


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