1、问题：

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 (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).

分析：

假设你有一个数组，其中第i个元素是一个给定股票第i天的价格。请找出最大利润值，可以交易多次，即多次购买和售出，但是在再次买进之前必须卖出之前买的。

       只要判断相邻元素是否递增即可，只要增加就把利润差累积起来。

2、解法：

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

注意数组的两端下标的临界值不要越界。