破解 LeetCode 389：巧用数据结构和异或巧解两数组差异

2023-12-04 15:56:40

在 LeetCode 389 的难题中，我们面临着一个看似简单的任务：找出两个字符串之间的差异字符。虽然这看起来像一个直接的比较问题，但 Python 提供了巧妙的方法，使用数据结构和异或运算来优雅高效地解决它。

1. 字典计数法

字典以键值对的形式存储数据，非常适合计数字符。我们的策略是遍历两个字符串，为每个字符创建一个字典，其中键是字符，值是其出现次数。然后，我们可以比较字典来找出差异字符。

def find_the_difference(s, t):
  """
  Finds the character that is in t but not in s.

  Args:
    s (str): The first string.
    t (str): The second string.

  Returns:
    str: The character that is in t but not in s.
  """

  # Create dictionaries for each string.
  s_dict = {}
  for char in s:
    if char not in s_dict:
      s_dict[char] = 0
    s_dict[char] += 1

  t_dict = {}
  for char in t:
    if char not in t_dict:
      t_dict[char] = 0
    t_dict[char] += 1

  # Find the character that is in t but not in s.
  for char in t:
    if char not in s_dict or s_dict[char] != t_dict[char]:
      return char

  # No difference found.
  return None

2. 异或运算法

异或（^) 运算符对两个二进制位进行操作。如果两个位相同，结果为 0；如果不同，结果为 1。我们可以利用异或的这一特性来解决我们的问题。

def find_the_difference(s, t):
  """
  Finds the character that is in t but not in s.

  Args:
    s (str): The first string.
    t (str): The second string.

  Returns:
    str: The character that is in t but not in s.
  """

  # Convert the strings to ASCII character codes.
  s_codes = [ord(char) for char in s]
  t_codes = [ord(char) for char in t]

  # Perform XOR operation on the character codes.
  result = 0
  for i in range(len(s_codes)):
    result ^= s_codes[i]
  for i in range(len(t_codes)):
    result ^= t_codes[i]

  # Convert the result back to a character.
  return chr(result)