股票买卖问题

最新推荐文章于 2024-05-12 13:48:18 发布

原创最新推荐文章于 2024-05-12 13:48:18 发布 · 193 阅读

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股票买卖问题

121、买卖股票的最佳时机
题目：
给定一个数组，它的第 i 个元素是一支给定股票第 i 天的价格。
如果你最多只允许完成一笔交易（即买入和卖出一支股票一次），设计一个算法来计算你所能获取的最大利润。
注意：你不能在买入股票前卖出股票

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if not prices: return 0
        n = len(prices)
        dp = [[0,0] for _ in range(n)]#创建n个二维数据[0,0],即[[0,0],[0,0],...,[0,0]]
        dp[-1][0] = 0
        dp[-1][1] = - float('inf')
        dp[0][0] = 0
        dp[0][1] = - float('inf')
        for i in range(n):
            dp[i][0] = max(dp[i-1][0],dp[i-1][1]+ prices[i]) 
            dp[i][1] = max(dp[i-1][1],-prices[i])
            #dp[i][1] = max(dp[i-1][1],dp[i-1][0]-prices[i])即i-1天没有拥有股票，即dp[i-1][0]=0
        return dp[-1][0]
```python


```python 另一种解法
class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        minprice = float('inf')
        maxProfit = 0
        for price in prices:
            minprice = min(minprice, price)
            maxProfit = max(maxProfit, price-minprice)
        return maxProfit

122、买卖股票的最佳时机 II

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if not prices: return 0
        n = len(prices)
        dp = [[0,0] for _ in range(n)]
        dp[-1][0] = 0
        dp[-1][1] = - float('inf')
        dp[0][0] = 0
        dp[0][1] = - float('inf')
        for i in range(n):
            dp[i][0] = max(dp[i-1][0],dp[i-1][1] + prices[i])
            dp[i][1] = max(dp[i-1][1],dp[i-1][0] - prices[i])
        return dp[-1][0]

另一种解法

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

123、买卖股票的最佳时机 III

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if not prices: return 0
        n = len(prices)
        dp = [[[0 for i in range(2)] for i in range(3)] for i in range(n)]
        for k in range(3):
            dp[0][k][1] = -float('inf')
            dp[-1][k][1] = -float('inf')
        for i in range(n):
            for k in range(1, 2+1):

                dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k][1]+prices[i])
                dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k-1][0]-prices[i])
                
        return dp[-1][2][0]

188、买卖股票的最佳时机 IV

class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        if not prices: return 0
        n = len(prices)
        if k >= n//2:
            res = 0
            for i in range(1,n):
                if prices[i] > prices[i-1]:
                    res += prices[i] - prices[i-1]
            return res
        else:
            dp = [[[0 for i in range(2)]  for _ in range(k+1)] for __ in range(n)]
            for t in range(k+1):
                dp[-1][t][1] = -float('inf')
                dp[0][t][1] = -float('inf')
            for i in range(n):
                for j in range(1,k+1):
                    dp[i][j][1] = max(dp[i-1][j][1],dp[i-1][j-1][0] - prices[i])
                    dp[i][j][0] = max(dp[i-1][j][0],dp[i-1][j][1] + prices[i])
            return dp[-1][-1][0]

309、买卖股票的最佳时机含冷冻期

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if not prices: return 0
        if len(prices) == 1: return 0
        n = len(prices)
        dp = [[0 for i in range(2)] for _ in range(n)]
        dp[-1][1] = -float('inf')
        dp[-2][1] = -float('inf')
        dp[0][1] = -float('inf')
        for i in range(n):
            dp[i][0] = max(dp[i-1][0],dp[i-1][1] + prices[i])
            dp[i][1] = max(dp[i-1][1],dp[i-2][0] - prices[i])
        return dp[-1][0]
    
另一种解法

```python
class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        n = len(prices)
        if n < 2: return 0
        dp = [[0,0] for i in range(n)]   #dp的初始化

        dp[0][0] = 0     #第0天不持股自然就为0了
        dp[0][1] = -prices[0]   #第0天持股，那么价格就是-prices[0]了
        #第1天不持股，要么第0天就不持股，要么就是第0天持股，然后第1天卖出
        dp[1][0] = max(dp[0][0],dp[0][1]+prices[1]) 
        #第一天持股，要么就是第0天就持股了，要么就是第0天不持股第1天持股
        dp[1][1] = max(dp[0][1],dp[0][0]-prices[1])
        
        for i in range(2,n):
            dp[i][0] = max(dp[i-1][0],dp[i-1][1]+prices[i])
            dp[i][1] = max(dp[i-1][1],dp[i-2][0]-prices[i])
        return dp[-1][0]


714、买卖股票的最佳时机含手续费

```python
class Solution:
    def maxProfit(self, prices: List[int], fee: int) -> int:
        if not prices: return 0
        n = len(prices)
        dp =[ [0,0] for _ in range(n)]
        dp[-1][1] = -float('inf')
        dp[0][1] = -float('inf')
        for i in range(n):
            dp[i][0] = max(dp[i-1][0],dp[i-1][1] + prices[i] -fee)
            dp[i][1] = max(dp[i-1][1],dp[i-1][0] - prices[i])
        return dp[-1][0]