PROBLEM DESCRIPTION
You are given an integer N and the task is to reverse the digits of the given integer. Return 0 if the result overflows and does not fit in a 32 bit signed integer.
SOLUTION
We can get the last digit using A%10. We can add it to a long variable res in reverse by using res = res*10 + (A%10). This will give us the reverse value. Since A can be negative, we can take a absolute value of it while reversing. To handle overflow, we can calculate the result as long and then check if it in between [Integer.MIN_VALUE, Integer.MAX_VALUE].
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public class Solution {
public int reverse(int A) {
long reverseA = 0;
int sign = (A < 0 ? -1 : 1);
A = Math.abs(A);
while(A != 0){
reverseA = reverseA*10 + (A%10);
A = A/10;
}
if(reverseA < Integer.MIN_VALUE || reverseA > Integer.MAX_VALUE)
return 0;
return sign * (int) reverseA;
}
}