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).
题意：121的followup，不限制交易次数了，问能获得的最大收益是多少。
思路：因为不限制交易次数了，所以只要相邻的两天有利润，我们就可以选择交易；如果当天的价格比前一天低，那么就不予考虑交易。是一种贪心的方法。
public int maxProfit(int[] prices) {
int profit = 0;
if (prices == null || prices.length < 2) {
return profit;
}
for (int i = 1; i < prices.length; i++) {
if (prices[i] > prices[i-1]) {
profit += prices[i] - prices[i-1];
}
}
return profit;
}