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[leetcode] Valid Palindrome @ Python
原题地址:https://oj.leetcode.com/problems/valid-palindrome/
题意:
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
"radar" is a palindrome
"rotator" 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.
解题思路1:利用python的list comprehension 以及string 的 isalnum()函数我们可以写出极为简短的解决方法 (原创):
class Solution: # @param s, a string # @return a boolean def isPalindrome(self, s): newS= [i.lower() for i in s if i.isalnum()] #return newS == newS[::-1] return newS[:len(newS)/2] == newS[(len(newS)+1)/2:][::-1]
需要说明的是程序的最后一行对于newS长度为奇或者为偶都通用
newS[:len(newS)/2] == newS[(len(newS)+1)/2:][::-1]
解题思路2:将不是字母的字符去掉,然后转换成小写,然后简单的回文判断。
class Solution: # @param s, a string # @return a boolean def isPalindrome(self, s): if s == ‘‘: return True else: sTmp = ‘‘ for i in range(0, len(s)): if s[i] >= ‘a‘ and s[i] <= ‘z‘ or s[i] >= ‘0‘ and s[i] <= ‘9‘ or s[i] >= ‘A‘ and s[i] <= ‘Z‘: sTmp += s[i] sTmp = sTmp.lower() for i in range(0, len(sTmp)/2): if sTmp[i] != sTmp[len(sTmp)-1-i]: return False return True
参考致谢:
[1]http://www.cnblogs.com/zuoyuan/p/3765882.html
[leetcode] Valid Palindrome @ Python