# My accepted solution in Java , which I think is really smart

• 1.char[] ary=s.toCharArray();

2.Start to travel ary

• If given parameter a is ''Valid Palindrome''
• Then for each element which is a alphanumeric character in ary
• There must be a same element at a correct position from the end of the ary
• So, If I couldn't find it , it would be fasle, or it will be true

``````public class Solution {
public boolean isPalindrome(String s) {
if(s.length()==0 || s.length()==1) return true;
int length=0;
char[] ary=s.toCharArray();
int j=ary.length-1;
for(int i=0;i<ary.length;i++)
{
if(('a'<=ary[i] && ary[i]<='z')||('A'<=ary[i] && ary[i]<='Z')||('0'<=ary[i] && ary[i]<='9'))
{
while(true)
{
if(j==-1) return false;
if(('a'<=ary[j] && ary[j]<='z')||('A'<=ary[j] && ary[j]<='Z')||('0'<=ary[j] && ary[j]<='9'))
{
if((ary[j]!=ary[i]) && (ary[j]+32!=ary[i]) && (ary[j]-32!=ary[i])) return false;
j--;
break;
}
j--;
}
}
}
return true;
}
}``````

• ``````def isPalindrome(s):
newS= [i.lower() for i in s if i.isalnum()]
return newS == newS[::-1]
``````

life is short, you need python

• This post is deleted!

• you're so smart... i'm shocked

• Hi man, you are so cool

• me too. hahaha

• WTF.....dude your code is killing me.

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