problem：
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).
Array Greedythinking：
其实就是求一个数组两个数之差的和最大值，采用贪心策略，有钱赚就不放过！！！收益就最大。
code：
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n=prices.size();
int profit=0;
for(int i=1;i<n;i++)
if(prices[i]-prices[i-1]>0)
profit+=prices[i]-prices[i-1];
return profit;
}
};leetcode || 122、Best Time to Buy and Sell Stock II
原文地址：http://blog.csdn.net/hustyangju/article/details/45306493
