1+ /*
2+ Author: Andy, nkuwjg@gmail.com
3+ Date: Aug 22, 2013
4+ Problem: Palindrome Number
5+ Difficulty: Easy
6+ Source: https://oj.leetcode.com/problems/palindrome-number/
7+ Notes:
8+ Determine whether an integer is a palindrome. Do this without extra space.
9+ Some hints:
10+ Could negative integers be palindromes? (ie, -1) (No!)
11+ If you are thinking of converting the integer to string, note the restriction of using extra space.
12+ You could also try reversing an integer. However, if you have solved the problem "Reverse Integer",
13+ you know that the reversed integer might overflow. How would you handle such case?
14+ There is a more generic way of solving this problem.
15+
16+ Solution: 1. Count the number of digits first (traverse once) then check the digits from both sides to center.
17+ 2. Reverse the number, then check to see if x == reverse(x).
18+ 3. Recursion (interesting but a little hard to understand). -> See C++.
19+ */
20+ public class Solution {
21+ public boolean isPalindrome (int x ) {
22+ return isPalindrome_2 (x );
23+ }
24+ public boolean isPalindrome_1 (int x ) {
25+ if (x <0 )return false ;
26+ int d =1 ;
27+ while (x /d >=10 )d *=10 ;
28+ while (d >1 ) {
29+ if (x %10 !=x /d )return false ;
30+ x = (x %d ) /10 ;
31+ d /=100 ;
32+ }
33+ return true ;
34+ }
35+ public boolean isPalindrome_2 (int x ) {
36+ if (x <0 )return false ;
37+ return x ==reverse (x );
38+ }
39+ public int reverse (int x ) {
40+ int res =0 ;
41+ while (x >0 ) {
42+ res =res *10 +x %10 ;
43+ x =x /10 ;
44+ }
45+ return res ;
46+ }
47+ }
