Easy

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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.

C++:

峰谷值捕捉：

/*
 * @Author: SourDumplings
 * @Link: https://github.com/SourDumplings/
 * @Email: [email protected]
 * @Description: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/
 * @Date: 2019-03-19 17:54:29
 * 
 * 想象一下，肯定是谷值处买，然后峰值处卖，这样最赚钱
 * 
 */

class Solution
{
  public:
    int maxProfit(vector<int> &prices)
    {
        if (prices.empty())
            return 0;
        int maxProfit = 0, n = prices.size();
        int peak = prices[0], valley = prices[0];
        int i = 0;
        while (i < n - 1)
        {
            while (i < n - 1 && prices[i + 1] <= prices[i])
                ++i;
            valley = prices[i];
            while (i < n - 1 && prices[i + 1] >= prices[i])
                ++i;
            peak = prices[i];
            maxProfit += peak - valley;
        }
        return maxProfit;
    }
};

One Pass：

/*
 * @Author: SourDumplings
 * @Link: https://github.com/SourDumplings/
 * @Email: [email protected]
 * @Description: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/
 * @Date: 2019-03-19 17:54:29
 * 
 * 就是一点一点往峰值上爬，只要股价涨了就卖，积小胜为大胜
 * 
 */

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

        return maxProfit;
    }
};