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.
public class Solution {
public boolean isPalindrome(String s) {
// Start typing your Java solution below
// DO NOT write main() function
if(s.isEmpty()) {
return true;
}
int left = 0;
int right = s.length()-1;
s = s.toLowerCase();
loop: while(left<right) {
while(!(s.charAt(left)>='a' && s.charAt(left)<='z') && !(s.charAt(left)>='0' && s.charAt(left)<='9')) {
left++;
if(left>=right) {
break loop;
}
}
while(!(s.charAt(right)>='a' && s.charAt(right)<='z') && !(s.charAt(right)>='0' && s.charAt(right)<='9')) {
right--;
if(right<=left) {
break;
}
}
if(s.charAt(left)!=s.charAt(right)) {
return false;
} else {
left++;
right--;
}
}
return true;
}
}
===========
public class Solution {
public boolean isPalindrome(String s) {
if (s == null || s.isEmpty()) {
return true;
}
int start = 0;
int end = s.length() - 1;
s = s.toLowerCase();
while (start <= end) {
while (start <= end && !Character.isLetterOrDigit(s.charAt(start))) {
start += 1;
}
while (start <= end && !Character.isLetterOrDigit(s.charAt(end))) {
end --;
}
if (start <= end) {
if (s.charAt(start) == s.charAt(end)) {
start += 1;
end -= 1;
} else {
return false;
}
}
}
return true;
}
}
没有评论:
发表评论