Description

Given a 32-bit 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

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

Submission

class Solution {
    public int reverse(int x) {
        int sign = 1;
        if(x < 0) {
            sign = -1;
            x = -x;
        }
        
        String sX = Integer.toString(x).trim();
        StringBuilder sb = new StringBuilder();
        for(int i = sX.length() - 1; i >= 0; i--) {
            sb.append(sX.charAt(i));
        }
        try {
            return sign * Integer.parseInt(sb.toString());
        } catch (Exception e) {
            return 0;
        }
    }
}

For small integers, it will work, but for larger ones, it won’t work if without the exception-handling part.

Solution

If we are to implement a library function, we should implement the function like the following.

class Solution {
    public int reverse(int x) {
        int rev = 0;
        while (x != 0) {
            int pop = x % 10;
            x /= 10;
            if (rev > Integer.MAX_VALUE/10 || (rev == Integer.MAX_VALUE / 10 && pop > 7)) return 0;
            if (rev < Integer.MIN_VALUE/10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) return 0;
            rev = rev * 10 + pop;
        }
        return rev;
    }
}