題目：
Determine whether an integer is a palindrome. Do this without extra space.
Could negative integers be palindromes? (ie, -1)
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.
題解：
最開始是用Reverse Integer的方法做的，通過了OJ，不過那個並沒有考慮越界溢位問題。遇見這種處理Number和Integer的問題，首要考慮的就是會不會溢位越界。然後再考慮下負數，0。還有就是要心裡熟悉\的結果和%的結果。感覺這種題就是考察你對數字敏感不敏感（反正我是很不敏感）
有缺陷的程式碼如下：
2 if(x<0)
3 return false;
4
5 int count = 0;
6 int testx = x;
7 while(testx!=0){
8 int r = testx%10;
9 testx = (testx-r)/10;
10 count++;
11 }
12
13 int newx = x;
14 int result = 0;
15 while(newx!=0){
16 int r = newx%10;
17 int carry = 1;
18 int times = count;
19 while(times>1){
20 carry = carry*10;
21 times--;
22 }
23 result = result+r*carry;
24 newx= (newx-r)/10;
25 count--;
26 }
27
28 return result==x;
29
30 }
另外一種方法可以避免造成溢位，就是直接安裝PalidromeString的方法，就直接判斷第一個和最後一個，迴圈往復。這樣就不會對數字進行修改，而只是判斷而已。
程式碼如下：
2 //negative numbers are not palindrome
3 if (x < 0)
4 return false;
5
6 // initialize how many zeros
7 int div = 1;
8 while (x / div >= 10) {
9 div *= 10;
10 }
11
12 while (x != 0) {
13 int left = x / div;
14 int right = x % 10;
15
16 if (left != right)
17 return false;
18
19 x = (x % div) / 10;
20 div /= 100;
21 }
22
23 return true;
24 }
Reference：
http://www.programcreek.com/2013/02/leetcode-palindrome-number-java/