股票买卖问题

class Solution(object):
    def colorBorder(self, grid, r0, c0, color):
        """
        :type grid: List[List[int]]
        :type r0: int
        :type c0: int
        :type color: int
        :rtype: List[List[int]]
        """
        old_c = grid[r0][c0]
        if old_c==color:return grid
        m = len(grid)
        n = len(grid[0])
        dp=[[0]*n for i in range(m)]
        def is_b(x,y):
            
            if x<0 or y<0 or x>=m or y>=n:
                return 1
            elif grid[x][y]!=old_c and dp[x][y]==0:
                return 1
            return 0
        def dfs(x,y):
            
            if x<0 or y<0 or x>=m or y>=n:
                return 
            elif grid[x][y]!=old_c:
                return 
            elif dp[x][y]==1:
                return 
            if is_b(x-1,y) or is_b(x+1,y) or is_b(x,y-1) or is_b(x,y+1):
                grid[x][y] = color 
            
            dp[x][y] =1  
            dfs(x-1,y)
            dfs(x+1,y)
            dfs(x,y-1)
            dfs(x,y+1)
        dfs(r0,c0)
        return grid 

　

股票利润：

1、只做一次交易：

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if len(prices)==0:
            return 0
        min1 = prices[0]
        n = len(prices)
        for i in range(n):
            min1 = min(min1,prices[i])
            prices[i] = prices[i]-min1
        return max(prices)

2、多次交易：

122. 买卖股票的最佳时机 II

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        min1 =10e4
        dp = 0

        for i in range(1,len(prices)):
            dp +=max(0,prices[i]-prices[i-1])
        return dp

3、只做2次交易：

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        
        n = len(prices)
        if n==0:return 0
        dp1=[0]*n
        dp2=[0]*n
        min1 = prices[0]
        for i in range(1,n):
            min1 =min(min1,prices[i])
            dp1[i] = prices[i]-min(min1,prices[i])
        max1 = prices[-1]
        for j in range(n-2,-1,-1):
            dp2[j] = max(max(prices[j],max1)-prices[j],dp2[j+1])
            max1 = max(prices[j],max1)
        max1 = dp1[-1]
        # print(dp1,dp2)
        for i in range(n-1):
            max1= max(max1,dp1[i]+dp2[i+1])
        return max1
 

含手续费：

714. 买卖股票的最佳时机含手续费

class Solution(object):
    def maxProfit(self, prices, fee):
        """
        :type prices: List[int]
        :type fee: int
        :rtype: int
        """

        sell = 0
        buy = -5e6
        for i in range(len(prices)):
            buy = max(sell-prices[i],buy)
            sell = max(buy+prices[i]-fee,sell)
        return sell

309. 最佳买卖股票时机含冷冻期

需要一个sell 记录前两天完成交易的利润，sell1表示前一天的；

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :type fee: int
        :rtype: int
        """

        sell = 0
        buy = -5e6
        if len(prices)<=1:return 0
        buy = max(sell-prices[0],buy)
        sell = max(buy+prices[0],sell)
        sell1 = max(buy+prices[1],sell)
        buy1 = max(-prices[1],buy)
       
        for i in range(2,len(prices)):
            tmp = sell1
            sell1 = max(buy1+prices[i],sell1)
            buy1 = max(sell-prices[i],buy1)
            sell  = tmp
            
        return sell1

 

特例：k>数组的一半长度    md

188. 买卖股票的最佳时机 IV

class Solution(object):
    def maxProfit(self, k, prices):
        """
        :type k: int
        :type prices: List[int]
        :rtype: int
        """

        n = len(prices)
        if n<=1:return 0
        if k>=n:
            max1 = prices[0]
            d = 0
            for i in range(1,len(prices)):
                if prices[i]>prices[i-1]:
                    d+=prices[i]-prices[i-1]
            return d
        dp=[[[0]*2 for j in range(k+1)] for i in range(n+1)]
        dp[0][0][1] = -prices[0]

        for i in range(n+1):#jiaoyi 1ci
            dp[i][0][0] = 0#max(dp[i-1][0][1]+prices[i],dp[i-1][0][0])
            dp[i][0][1] = -5e6#max(dp[i-1][0][0]-prices[i],dp[i-1][0][1])
        for j in range(k+1):
            dp[0][j][0] = 0
            dp[0][j][1] = -5e6
        
        for i in range(1,n+1):
            for j in range(1,k+1):    
                    dp[i][j][0] = max(dp[i-1][j][0],dp[i-1][j][1]+prices[i-1])
                    dp[i][j][1] = max(dp[i-1][j][1],dp[i-1][j-1][0]-prices[i-1])

        return dp[-1][k][0]