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 at most two transactions. 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: [3,3,5,0,0,3,1,4] Output: 6 Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 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.
class Solution(object): def maxProfit(self, prices): """ :type prices: List[int] :rtype: int """ # 这个规定最多交易两次，即两次买， 两次卖 # local[i][j] 代表在最多进行j次交易的情况下，并且在第i天当天卖掉之后的最大利润 # global[i][j] 代表在最多进行j次交易的情况下，第i天结束时的最大利润 # 则有 l[i][j] = max(g[i-1][j-1] + max(diff, 0), l[i-1][j]+diff) # g[i][j] = max(l[i][j], g[i-1][j]) if (len(prices)) ==0: return 0 n = len(prices) l = [[0 for _ in range(3)] for _ in range(n)] # n行3列 g = [[0 for _ in range(3)] for _ in range(n)] # n行3列 for i in range(1, len(prices)): diff = prices[i]- prices[i-1] # 昨天买今天卖的差值 for j in range(1, 3): l[i][j] = max(g[i-1][j-1]+max(diff, 0), l[i-1][j]+diff) # 因为有两种情况，首先今天必段卖，所以后面的，另外是如是要diff>0的话，如果diff<0的话，就不记了，因为可能还是之前的最大 # 注意虽然今天要卖，但是l代表的是前面这几天中最大的利润 g[i][j] = max(l[i][j], g[i-1][j]) # 更新之前的最大和当前的最大 return g[n-1]