# Accepted JAVA solution, none-recursion, No extra space

• This solution doesn't reverse the original integer, so we don't have to worry about the Integer.MAX_VALUE that was pointed out in the hint.

``````    if (x<0) return false;

int length = String.valueOf(x).length();

while(length>0){
int right = (int) (x/(Math.pow(10,(length-1))));
int left = x%10;
if( right != left) return false;
x = (int) (x%Math.pow(10,(length-1))); //remove the left most digit
x = x/10; //remove the right most digit
length = length - 2;
}
return true;
``````

It can also be written as following to avoid multiple computation of Pow(), but it's less intuitive:

``````    if (x<0) return false;

int length = String.valueOf(x).length();
int div = (int) (Math.pow(10,(length-1)));

while(div>0){
int right = x/div;
int left = x%10;
if( right != left) return false;
x = (x%div); //remove the left most bit
x = x/10; //remove the right most bit
div /= 100;
}
return true;``````

• This post is deleted!

• “int length = String.valueOf(x).length(); ”: Will converting the Integer to String lead to the extra space? But this solution will not cause overflow

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