Question:

Say you have an array for which the i 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).

假设您有一个数组，第i个元素是i天给定股票的价格。

设计了一种求最大利润的算法。您可以完成任意多的交易（即，多次买入和卖出股票的一部分）。

注：您不能同时进行多笔交易（即，您必须在再次购买之前卖出股票）。

Example:

  1. Input: [7,1,5,3,6,4]
  2. Output: 7
  3. Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
  4.              Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
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.

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Solution:

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {

    let total = 0;

    for(let i = 1; i < prices.length; i++){

        if(prices[i] - prices[i-1] > 0)
        total +=  prices[i] - prices[i-1]

    }
    return total

};