题目

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.

思路

“炒股”问题，要想利润最大化，那么就得低买高卖，赚取利润，如何判断有利润？只有当这一天的价格大于前一天的价格时才会有利润。可以遍历一趟整个数组，统计所有交易的利润，也就是把所有的正差价加起来。只要后一天的数字大于前一天，那么就把这两天之间的“利润”算进去。
因为所有的“涨势”，例如[1,2,3,4]，看起来只做了一组交易（1买4卖，profit=4-1=3），但其实每一天都在产生利润（profit= (2-1) + (3-2) + (4-3) =3）
时间复杂度 O(n)

代码

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

        }
        return profit;
    }
}

Runtime: 1 ms, faster than 88.49% of Java online submissions for Best Time to Buy and Sell Stock II.
Memory Usage: 35.6 MB, less than 95.92% of Java online submissions for Best Time to Buy and Sell Stock II.