# LeetCode题目：Palindrome Number

Palindrome Number
Determine whether an integer is a palindrome. Do this without extra space.
Some hints:
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.

```class Solution {
public:
bool isPalindrome(int x) {
//cout<<sizeof(long long);
if(x < 0)
return false;
long long xx = x;
long long s = 1;
while(xx / s)
s *= 10;
s /= 10;
long long e = 10;
//if(s < e) cout<<s<<"vs"<<e<<"forx"<<x;
while(s >= e) {
long long sd = (xx / s);
long long ed = (xx % e) / (e / 10);
if(sd!=ed) {
//cout<<sd<<"vs"<<ed<<"at"<<s;
return false;
}
//cout<<sd<<"vs"<<ed<<"at"<<s;
xx = xx % s;
s /= 10;
e *= 10;
}
return true;
}
};
```

## “LeetCode题目：Palindrome Number”的5个回复

1. baltimost说道：

请问如何处理10000021 类似 test case的问题呢？

1. 就按照算法描述中的从开头和结尾各取一位来比较，然后逐步往中间比较。

2. Shuanglong Zhang说道：

但是你还是用了好几个局部变量，那些不是extra space 吗

1. 有道理，我理解的extra space是堆上的空间，不过如果把局部变量也算上的话，我想这个题也是可以做的，可以用计算来替换这些局部变量。