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
class Solution { public boolean isPalindrome(String s) { if(s == null || s.length() == 0 || s.length() == 1){ return true; } int left = 0; int right = s.length() - 1; s = s.toLowerCase(); while(left < right){ if(!Character.isLetterOrDigit(s.charAt(left))){ left++; }else if(!Character.isLetterOrDigit(s.charAt(right))){ right--; }else if(s.charAt(left) != s.charAt(right)){ return false; }else{ left++; right--; } } return true; } }
below solution is written by me previously. It would work for 99.9% cases, except for one that failed , which is a very very very very long string.
class Solution {
public boolean isPalindrome(String s) {
if(s ==null||s.trim().length()==0){
return true;
}
String tmp = "";
for(int i = 0; i<s.length(); i++){
if( ((int) s.charAt(i) >=65 && (int)s.charAt(i)<=90)||((int) s.charAt(i) >=97 && (int)s.charAt(i)<=122)||((int) s.charAt(i) >=48 && (int)s.charAt(i)<=57))
tmp = tmp+s.charAt(i);
}
System.out.println(tmp.trim());
return Check( tmp.toLowerCase());
}
public boolean Check(String s){
int start = 0;
int end = s.length()-1;
for(int i = start, j= end; i<j; i++, j--){
if(s.charAt(i)!= s.charAt(j)){
return false;
}
}
return true;
}
}
(Easy) Valid Palindrome -LeetCode
原文：https://www.cnblogs.com/codingyangmao/p/11425672.html