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

题目地址

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/

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.

前置知识

• 数组

公司

• 阿里
• 腾讯
• 百度
• 字节
• bloomberg

思路

由于我们是想获取到最大的利润，我们的策略应该是低点买入，高点卖出。

由于题目对于交易次数没有限制，因此只要能够赚钱的机会我们都不应该放过。

如下图，我们只需要求出加粗部分的总和即可

用图表示的话就是这样：

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

语言支持：JS，C++，Java

JS Code:

/**
 * @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;
};

C++ Code:

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

Java Code:

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

复杂度分析

• 时间复杂度：$O(N)$
• 空间复杂度：$O(1)$

相关题目

• 121.best-time-to-buy-and-sell-stock
• 309.best-time-to-buy-and-sell-stock-with-cooldown

更多题解可以访问我的LeetCode题解仓库：https://github.com/azl397985856/leetcode 。 目前已经35K star啦。

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