LeetCode | 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 (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).

要求可以多次买入、卖出股票，但同一时间手中只能持有一只股票。

tags提示：greedy，提示用贪心算法实现
贪心算法：在对问题求解时，总是做出在当前看来最好的决策。其不从整体最优上加以考虑，所做出的仅是某种意义的局部最优解。

//greedy: 在第i天，根据第i+1天的价格来决策：
//如果第i+1天涨，在第i天就不卖，留到第i+1天做决策；如果第i+1天跌，在第i天就卖掉，然后在第i+1天再买入伺机卖出，如此就能获得所有上涨阶段所产生的收益
//实际实现时：类似于求所有价格上涨阶段的总涨幅
public class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length <= 1) return 0;
        
        int maxProfit = 0;
        for(int i=1; i<prices.length; i++){
            if(prices[i]>prices[i-1]){   
                maxProfit += prices[i]-prices[i-1];  //所有上涨阶段的收益总和
            }
        }
        return maxProfit;
    }
}