leetcode 122. Best Time to Buy and Sell Stock II

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本文介绍了一种算法，用于计算给定股票价格序列时的最大可能利润。算法允许进行多次买入和卖出操作，前提是每次卖出后才能再次买入。通过实例演示了如何在一系列股票价格中找到最佳交易时机。

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.

给出每天的股价，可以交易无数次，条件是卖了才能买入下一支股票

思路：
121， 123， 188买卖股票系列，
本题的greedy方法在188题中k > length/2时用到
就是只要有收益，就做交易
因为price[i] - price[j]
= (price[i] - price[i - 1]) + (price[i - 1] - price[i -2] + ,…, + (prices[j + 1] - prices[j])

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