## 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.
``````

## 思路

C++ 解法：

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

Java 解法：

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

## 关键点解析

• 这类题只要你在心中（或者别的地方）画出上面这种图就很容易解决

## 代码

JS Code:

``````
/*
* @lc app=leetcode id=122 lang=javascript
*
* [122] Best Time to Buy and Sell Stock II
*
*
* 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}
*/
var maxProfit = function(prices) {
let profit = 0;

for(let i = 1; i < prices.length; i++) {
if (prices[i] > prices[i -1]) {
profit  = profit + prices[i] - prices[i - 1];
}
}

return profit;
};
``````

Python Code:

``````class Solution:
def maxProfit(self, prices: 'List[int]') -> int:
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]))
``````

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