122. 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 (i.e., buy one and sell one share of the stock multiple times).

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

greedy，可以想象每天买卖（前一天买后一天卖）。如果某一天的利润>0，可以认为这天买入了股票并卖出得到利润，累加到利润和上；如果这天的利润<0，可以认为这天不买股票，不累加到利润和上。

detail: 用一个常量profit表示利润，初始为0。遍历整个数组，如果两数之差 > 0，累加到profit上。

时间：O(N)，空间：O(1)

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