题目:
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 (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).
思路:
1.目标是累加数组递增部分的差值
2.只需判断数组相邻两元素值的差值是否大于0,然后累加所有大于的0的差值
public class Solution { public int maxProfit(int[] prices) { int len=prices.length; //数组长度 int res=0; //最后结果 for(int i=1;i<len;i++) { if(prices[i]>prices[i-1]) res+=prices[i]-prices[i-1]; } return res; } }