122. Best Time to Buy and Sell Stock II

https://leetcode.com/problems/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 (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.

最大收益不错过任何低买高卖。直观的想法是找出每个谷和后面的峰并相减。但实际它俩的差距由它俩之间每两天的差别和构成。

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

或者DP:

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        N = len(prices)
        if N == 0:
            return 0
        buy, sell = -prices[0], 0
        for i in range(1, N):
            buy = max(buy, sell - prices[i])
            sell = max(sell, buy + prices[i])
        return sell