LeetCode刷题记录-20181021
119. Pascal's Triangle II
返回杨辉三角第n层数字列表,和118类似
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1 class Solution: 2 def getRow(self, rowIndex): 3 """ 4 :type rowIndex: int 5 :rtype: List[int] 6 """ 7 if rowIndex== 0: 8 return [1] 9 else: 10 result = [] 11 temp = self.getRow(rowIndex-1) 12 temp = [0] + temp 13 temp.append(0) 14 for i in range(len(temp)-1): 15 result.append(temp[i]+temp[i+1]) 16 return result
121. Best Time to Buy and Sell Stock
O(n)时间复杂度,列表中最小值为买点,依次迭代列表,找到最大利润。注意:只能买卖一次
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1 class Solution: 2 def maxProfit(self, prices): 3 """ 4 :type prices: List[int] 5 :rtype: int 6 """ 7 profit = 0 8 m = len(prices) 9 if m==0: 10 return 0 11 local_profit = 0 12 buypoint = prices[0] 13 for i in range(1,m): 14 if prices[i]<buypoint: 15 buypoint = prices[i] 16 local_profit = prices[i]-buypoint 17 if local_profit>profit: 18 profit = local_profit 19 return profit
122.Best Time to Buy and Sell Stock II
可以买卖多次,卖完之后才可以买入。最大利润为所有差值为正的和
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1 class Solution: 2 def maxProfit(self, prices): 3 """ 4 :type prices: List[int] 5 :rtype: int 6 """ 7 sum = 0 8 m = len(prices) 9 for i in range(m-1): 10 temp = prices[i+1]-prices[i] 11 if temp >0: 12 sum += temp 13 return sum 14
125.Valid Palindrome
判断是否为回文(字符串包含空格、标点符号),判断字符是否为字符串或者数字,a.isalnum()
方法1:设定两个指针,头指正,尾指针,判断当前位置是否满足字符或者数字,如满足,判断头尾是否相同(注意大小写);如不满足,头指正加1,尾指针减一,O(n)时间复杂度
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1 class Solution: 2 def isPalindrome(self, s): 3 """ 4 :type s: str 5 :rtype: bool 6 """ 7 m = len(s) 8 if m == 0: 9 return True 10 left_tag = 0 11 right_tag = m -1 12 13 while left_tag <= right_tag: 14 left_letter = s[left_tag] 15 right_letter = s[right_tag] 16 if not left_letter.isalnum(): 17 left_tag+=1 18 continue 19 if not right_letter.isalnum(): 20 right_tag-=1 21 continue 22 23 if left_letter.upper() != right_letter.upper(): 24 return False 25 else: 26 left_tag +=1 27 right_tag -=1 28 29 return True
方法2:正则表达式,剔除非数字和字母字符,翻转后对比
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1 class Solution: 2 def isPalindrome(self, s): 3 """ 4 :type s: str 5 :rtype: bool 6 """ 7 s = re.sub('[^a-zA-Z0-9]', '', s).lower() 8 return s[::-1] == s