[LeetCode] 122. Best Time to Buy and Sell Stock II_Easy tag: Dynamic Programming

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.


思路为Dynamic Programming, 感觉实际上跟sell and buy在同一天实际上一样的结果? dp[i]表示目前最大的profit
动态方程式: dp[i] = dp[i-1] + max(0, prices[i] - prices[i-1])
init: dp[0] = 0

1. Constraints
1) edge case [] => 0

2. Ideas

Dynamic Programming     T: O(n)   S: O(1) using rolling array

3. Code

class Solution:
    def buySellStock2(self, prices):
        if not prices: return 0
        dp, n = [0]*2, len(prices)
        for i in range(1, n):
            dp[i%2] = dp[i%2-1] + max(0, prices[i] - prices[i-1])
        return dp[(n-1)%2]  # n-1因为我们要for loop里面最后的元素