122. 买卖股票的最佳时机II

1. 暴力法

   计算所有可能的交易组合相对应的利润，并找出它们其中的最大利润。

   • Python3

     class Solution:
         def maxProfit(self, prices: List[int]) -> int:
             return self.calculate(prices, 0)
     
         def calculate(self, prices, s):
             if s >= len(prices):
                 return 0
     
             max = 0
             for start in range(s, len(prices)):
                 max_profit = 0
                 for i in range(start+1, len(prices)):
                     if prices[start] < prices[i]:
                         profit = self.calculate(prices, i+1) + prices[i] - prices[start]
                         if profit > max_profit:
                             max_profit = profit
                 if max_profit > max:
                     max = max_profit
             return max
     

   • Go

     func maxProfit(prices []int) int {
         return cal(prices, 0)
     }
     
     func cal(prices []int, s int) int {
         if s > len(prices) {
             return 0
         }
         max := 0
         for i := s; i < len(prices); i++ {
             maxprices := 0
             for j := i+1; j < len(prices); j++ {
                 if prices[i] < prices[j]{
                     pro := (prices[j] - prices[i]) + cal(prices, j+1)
                     if pro > maxprices {
                         maxprices = pro
                     }
                 }
     
             }
             if max < maxprices {
                 max = maxprices
             }
         }
         return max
     }
     

2. 贪心法

   

   • 从第 i 天（这里 i >= 1）开始，与第 i - 1 的股价进行比较，如果股价有上升（严格上升），就将升高的股价（ prices[i] - prices[i- 1] ）记入总利润，按照这种算法，得到的结果就是符合题意的最大利润。

   • Python

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

   • Go

     func maxProfit(prices []int) int {
         max := 0
         for i :=1; i<len(prices); i++{
             if prices[i] > prices[i-1]{
                 max += prices[i] - prices[i-1]
             }
         }
         return max
     }