Given a 32-bit signed integer, reverse digits of an integer.

Note: Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For this problem, assume that your function returns 0 when the reversed integer overflows.

Example 1:

1 2	Input: x = 123 Output: 321

Example 2:

1 2	Input: x = -123 Output: -321

Example 3:

1 2	Input: x = 120 Output: 21

Example 4:

1 2	Input: x = 0 Output: 0

Constraints:

1	-2ˆ31 <= x <= 231 - 1

Java

x /= 10;

x = 123456789;
x = 12345678;
…

x % 10;
9
8
…

reverse = reverse * 10 + x % 10;
987
…

(int)reverse==reverse?(int)reverse:0;
check integer overflow -2,147,483,647 to 2,147,483,647

import java.util.*;
class Solution {
    public int reverse(int x) {
        long reverse = 0;  
        while(x != 0){  
            reverse = reverse * 10 + x % 10;  // construct reverse integer
            x /= 10;  // get integer except last
        }  
        return (int)reverse==reverse?(int)reverse:0;
    }
}

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Lunski's Clutter

7. Reverse Integer

Java