LC Python 刷题笔记 有志者事竟成

楼主加油 不过也不用刷一千题。我四个月4000次submission tc 140-》280给楼主参考一下

老哥 还在刷吗？

1. '''
   
2. 
   
3. 6.    给一个数字列表且某些数字重复，给出每个数字还需要加进多少个才能使得列表里每个数字都一样多。给一个数字列表且某些数字重复，给出每个数字还需要加进多少个才能使得列表里每个数字都一样多。
   
4. 
   
5. '''
   
6. 
   
7. nums = [1, 3, 4, 1, 7, 8, 3, 2, 4, 5, 7, 2, 3, 5, 6, 4, 2, 8, 5]
   
8. 
   
9. output = {1: 1, 3: 0, 4: 0, 7: 1, 8: 1, 2: 0, 5: 0, 6: 2}
   
10. 
    
11. from collections import Counter
    
12. 
    
13. def helper(nums):
    
14.     dic = Counter(nums)
    
15.     max_num = sorted(list(dic.items()),key = lambda x:x[1], reverse = True)[0][1]
    
16.     dic1 = {}
    
17.     for c in dic:
    
18.         dic1[c] = max_num - dic[c]
    
19.     return dic1
    
20. 
    
21. print(helper(nums))

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1. '''
   
2. 2.    valid parenthesis or bracket
   
3. (Valid anagram)
   
4. (注意比lc上多一个<>括号，我一开始也没注意 出了bug才发现)
   
5. (valid 括号:（）<> {} [])
   
6. 
   
7. Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
   
8. 
   
9. An input string is valid if:
   
10. 
    
11. Open brackets must be closed by the same type of brackets.
    
12. Open brackets must be closed in the correct order.
    
13. Note that an empty string is also considered valid.
    
14. 
    
15. Example 1:
    
16. 
    
17. Input: "()"
    
18. Output: true
    
19. Example 2:
    
20. 
    
21. Input: "()[]{}"
    
22. Output: true
    
23. Example 3:
    
24. 
    
25. Input: "(]"
    
26. Output: false
    
27. Example 4:
    
28. 
    
29. Input: "([)]"
    
30. Output: false
    
31. Example 5:
    
32. 
    
33. Input: "{[]}"
    
34. Output: true
    
35. 
    
36. '''
    
37. 
    
38. def isValid(s):
    
39.     while '()' in s or '{}' in s or '[]' in s :
    
40.         s = s.replace('()','')
    
41.         s = s.replace('{}','')
    
42.         s = s.replace('[]','')
    
43.         s = s.replace('<>','')
    
44.     return s == ''

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1. def isValid(self, s: str) -> bool:
   
2.         brackets = ["()", "[]", "{}","<>"]
   
3.         while any(x in s for x in brackets):
   
4.             for bracket in brackets:
   
5.                 s = s.replace(bracket,'')
   
6.         return not s

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1. stack = []
   
2.         mapping = {")": "(", "}": "{", "]": "["}
   
3.         for char in s:
   
4.             if char in mapping:
   
5.                 top_element = stack.pop() if stack else '#'
   
6.                 if mapping[char] != top_element:
   
7.                     return False
   
8.             else:
   
9.                 stack.append(char)
   
10.         return not stack

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