LeetCode 122. Best Time to Buy and Sell Stock II

Say you have an array prices  for which the i^th 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.

Constraints:

1 <= prices.length <= 3*10^4

0<= prices[i] <= 10^4

解析：

此题目是在 LeetCode 121. Best Time to Buy and Sell Stock基础上进行的升级，即可以进行多次买入和卖出操作。根据题目要求，选择贪心法，从第二天开始，每天如果比前一天价格高则把差值加入利润中，因为我们可以昨天买入，今日卖出，若明日价更高的话，还可以今日买入，明日再抛出。以此类推，遍历完整个数组后即可求得最大利润。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int max_profit = 0;
        for(int i = 1; i < prices.size(); i++)
        {
            if(prices[i] > prices[i-1])
                max_profit += prices[i] - prices[i-1];
        }
        return max_profit;
    }
};