9 Palindrome Number(判断是否为回文数Easy)

class Solution {
public:
bool isPalindrome(int x) {
if(x<0)return false;
if(x==0)return true;
int start,end,num=0,temp=x;
while(temp){
num++;
temp=temp/10;
}
while(x){
start=x/int(pow(10,num-1));
end=x%10;
if(start!=end)return false;
x=x%int(pow(10,num-1))/10;
num-=2;
}
return true;
}
};