# 122. Best Time to Buy and Sell Stock II 买卖股票的最佳时机 II

@TOC

## # 题目描述

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.
``````

## # 解题方法

Java解法如下：

``````public class Solution {
public int maxProfit(int[] prices) {
int len = prices.length;
int ans = 0;
for(int i = 0; i < len - 1; i++){
if(prices[i + 1] > prices[i]){
ans += prices[i + 1] - prices[i];
}
}
return ans;
}
}
``````

``````class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
if not prices: return 0
ans = 0
for i in range(len(prices) - 1):
if prices[i + 1] > prices[i]:
ans += prices[i + 1] - prices[i]
return ans
``````

``````class Solution {
public:
int maxProfit(vector<int>& prices) {
int N = prices.size();
int res = 0;
for (int i = 1; i < N; i ++) {
if (prices[i] > prices[i - 1]) {
res += prices[i] - prices[i - 1];
}
}
return res;
}
};
``````

## # 日期

2017 年 4 月 20 日 2018 年 4 月 10 日 2018 年 11 月 26 日 —— 11月最后一周！