Leet Code OJ 125. Valid Palindrome [Difficulty: Easy] -python

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题目：

125. Valid Palindrome

Easy

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Note: For the purpose of this problem, we define empty string as valid palindrome.

Example 1:

Input: "A man, a plan, a canal: Panama"
Output: true

Example 2:

Input: "race a car"
Output: false

 

Constraints:

• s consists only of printable ASCII characters.

 题意：

题目意思是判断输入字符串是否是回文串，只判断字符串中的数字和字母；并且空字符串也算回文字符串。

思路：

1.过滤字符串，用正则表达式\W过滤非数字和非字母；

2.判断字符串左右是否相等

代码：

#125 Valid Palindrome
class Solution(object):def isPalindrome(self, s):""":type s: str:rtype: bool"""if s=='' :return Trues1 = s.lower()s2 = re.sub('\W+','',s1).replace('_','')
#         print(s2)l = len(s2)s3 = s2[::-1] #颠倒字符串if s2[0:int(l/2)]==s3[0:int(l/2)]:return Trueelse:return False

或者直接不颠倒，直接反向读取：这里注意反向读取字符串的写法：s2[:-(int(l/2)+1):-1]，自己试了一下可以这样反向截取。

#125 Valid Palindrome
class Solution(object):def isPalindrome(self, s):""":type s: str:rtype: bool"""if s=='' :return Trues1 = s.lower()s2 = re.sub('\W+','',s1).replace('_','')
#         print(s2)l = len(s2)print(int(l/2))
#         s3 = s2[::-1]
#         print(s2[0:int(l/2)],s2[:-(int(l/2)+1):-1])if s2[0:int(l/2)]==s2[:-(int(l/2)+1):-1]:return Trueelse:return False

 

 

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