eric111116
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+/*
+ Author:     Andy, nkuwjg@gmail.com
+ Date:       Aug 22, 2013
+ Problem:    Palindrome Number
+ Difficulty: Easy
+ Source:     https://oj.leetcode.com/problems/palindrome-number/
+ Notes:
+ Determine whether an integer is a palindrome. Do this without extra space.
+ Some hints:
+ Could negative integers be palindromes? (ie, -1) (No!)
+ If you are thinking of converting the integer to string, note the restriction of using extra space.
+ You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", 
+ you know that the reversed integer might overflow. How would you handle such case?
+ There is a more generic way of solving this problem.
+
+ Solution: 1. Count the number of digits first (traverse once) then check the digits from both sides to center.
+           2. Reverse the number, then check to see if x == reverse(x).
+           3. Recursion (interesting but a little hard to understand). -> See C++.
+ */
+public class Solution {
+    public boolean isPalindrome(int x) {
+        return isPalindrome_2(x);
+    }
+    public boolean isPalindrome_1(int x) {
+        if (x < 0) return false;
+        int d = 1;
+        while (x / d >= 10) d *= 10;
+        while (d > 1) {
+            if (x % 10 != x / d) return false;
+            x = (x % d) / 10;
+            d /= 100;
+        }
+        return true;
+    }
+    public boolean isPalindrome_2(int x) {
+        if (x < 0) return false;
+        return x == reverse(x);
+    }
+    public int reverse(int x) {
+        int res = 0;
+        while (x > 0) {
+            res = res * 10 + x % 10;
+            x = x / 10;
+        }
+        return res;
+    }
+}