# LeetCode | Palindrome Number（回文数字）

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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (x < 0)
return false;
if (x == 0)
return true;

int base = 1;
while(x / base >= 10)
base *= 10;

while(x)
{
int leftDigit = x / base;
int rightDigit = x % 10;
if (leftDigit != rightDigit)
return false;

x -= base * leftDigit;
base /= 100;
x /= 10;
}

return true;
}
};``````

``````class Solution {
public:
bool isPalindrome(int x) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(x < 0) return false;

long long target = x;
//if(target < 0) target = -target;
long long reverse = 0;
long long curNum = target;
while(curNum != 0)
{
reverse = 10*reverse+curNum%10;
curNum /= 10;
}

return reverse == target;
}
};``````

``````#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int ReverseInteger(int n);

int main(void)
{
int n;

while(scanf("%d",&n) == 1){
if(flag)
printf("true\n");
else
printf("false\n");
}

return 0;
}

{
unsigned int temp;
int numhi,numlo;
int pownum;

if(n == 0)
return 1;

if(n < 0){  //将负数转化成正数判断
temp = -n;
}else
temp = n;

int Nbit = 0;

while(n){
++Nbit;
n /= 10;
}

if(Nbit % 2){
Nbit = Nbit/2;
pownum = (int)pow(10,Nbit);
numlo = temp % pownum;
numhi = temp / (pownum*10);
}else{
Nbit = Nbit/2;
pownum = (int)pow(10,Nbit);
numlo = temp % pownum;
numhi = temp / pownum;
}

numlo = ReverseInteger(numlo);
printf("numhi = %d,numlo = %d\n",numhi,numlo);
if(numhi == numlo)
return 1;
return 0;
}

int ReverseInteger(int n)
{
if(n == 0)
return 0;
unsigned int temp;
if(n < 0)
temp = -n;
else
temp = n;

int result = 0,remainder;
while(temp){
remainder = temp % 10;
result = result * 10 + remainder;
temp /= 10;
}

return result;
}
``````