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：买卖股票的最佳时机，但这次能够完成任意数量的交易，，但不能同时进行多项交易，即在买之前必须先卖。
解决思路：
代码：能赚就买卖……
{(int[] prices) {int max = 0;for(int i = 0; i < prices.length – 1; i++){if(prices[i + 1] > prices[i]){max += prices[i + 1] – prices[i];}}return max;}}
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