LeetCode Valid Palindrome
Problem
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
For example, “A man, a plan, a canal: Panama” is a palindrome. “race a car” is not a palindrome.
Note: Have you consider that the string might be empty? This is a good question to ask during an interview.
For the purpose of this problem, we define empty string as valid palindrome.
即判断一个字符串的字符部分是不是回环字符串。即出去非字母，数字以外的字符，剩下的部分满足正反相同。
Python实现
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'''
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
For example,
"A man, a plan, a canal: Panama" is a palindrome.
"race a car" is not a palindrome.
Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.
For the purpose of this problem, we define empty string as valid palindrome.
'''
# author li.hzh
class Solution:
def isPalindrome(self, s):
"""
:type s: str
:rtype: bool
"""
if len(s) == 0:
return True
front_index, end_index = 0, len(s) - 1
while front_index < end_index:
if not s[front_index].isalnum():
front_index += 1
continue
if not s[end_index].isalnum():
end_index -= 1
continue
if s[front_index].lower() != s[end_index].lower():
return False
else:
front_index += 1
end_index -= 1
return True
print(Solution().isPalindrome("A man, a plan, a canal: Panama"))
print(Solution().isPalindrome("race a car"))
分析
很直接的思路，首尾两个指针，依次过滤字符进行判断。有不相等即返回False。
还有一个思路，代码很简单。先用正则去掉所有的非字母数字的字符，然后判断原字符串与逆序的字符串相等即可。该思路用python的re包实现非常简单。
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class Solution:
def isPalindrome(self, s):
"""
:type s: str
:rtype: bool
"""
import re
s = re.sub(r'[^A-Za-z0-9]', '', s).lower()
return s == s[::-1]
所有代码已上传至Github：https://github.com/lihongzheshuai/yummy-code
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