Problem Statement
Given a signed integer, reverse digits of an integer.
Example 1:
Input: 123
Output: 321
Example 2:
Input: 123
Output: 321
Example 3:
Input: 120
Output: 21
Return 0 for Integer overflow
Algorithm
The algorithm should be simple. You fetch the last digit. For next round, you need to move this to next digit place, and add next last digit.
 Keep a flag indicating if the number is negative. Just to negate the result while returning And, make the number positive if it is negative
 You need to get last digit from number, and multiply last result by 10 to move it to next place.
 You need to take care of the integer overflow.
 We will do two calculations
 Multiply last remainder result by 10
 Add current remainder with above multiplication result

Overflow cases: Note, max limit for 64bit integer: 2^(641)  1 = 2147483647

case 1: Multiplication is itself overflowed: Operation we will do: s10, we can check if s10 > Integer.MAX_VALUE OR, s > Integer.MAX_VALUE/10

case 2: if s10 is equal to Integer.MAX_VALUE, Which means the number will be *214748364 multiply it with 10 will give: 21474836470 So, we need to check the remainder (which is to be added), if it is greater than 7, it is an overflow.

Code
public class Q7_ReverseInteger {
public int reverse(int x) {
boolean neg = false;
if (x < 0) {
//negative number
neg = true;
x = x;
}
int s = 0;
while (x > 0) {
int rem = x%10;
if (s > Integer.MAX_VALUE/10  (s == Integer.MAX_VALUE/10 && rem > 7)) {
return 0;
}
s = s*10 + rem;
x = x/10;
}
if (neg) {
return s;
}
return s;
}
}