Best Time to Buy and Sell Stock II

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).

与之前不同的是可以多次买卖。我的思路是当前数比前一个小时，就卖出前面的。

public class Solution {    public int maxProfit(int[] prices) {        if(prices==null||prices.length<=0)return 0;        int minElement=prices[0],dif=0,t=0;for(int i=1;i<prices.length;i++){if(prices[i-1]>prices[i]){minElement=prices[i];dif+=t;t=0;continue;}else{t=Math.max(t, prices[i]-minElement);minElement=Math.min(minElement, prices[i]);}}return dif+t;    }}

更简单的思路是，数组递增就计入利润，否则继续。

public int maxProfit3(int[] prices) {    int profit = 0;    for(int i=1; i<prices.length; i++){        int diff = prices[i]-prices[i-1];        if(diff > 0){            profit += diff;        }    }    return profit;}