LeetCode 7. Reverse Integer

Description

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

Analysis

Two solutions are shown.
The first one may be the most average one. To reduce run time, I delete the zeroes at the end (except 0 itself) and terminate the while loop early when the result overflows.
The second is to use string.
The former performs better on the run time as to the test cases.
The key is how to cope with overflow. A convenient way is to store the result in a long variable.

Code

Version 1 Math

class Solution {public:    int reverse(int x) {        if (x == 0) return 0;        long result = 0;        while (x % 10 == 0){            x /= 10;        }        while (x){            result = result * 10 + x % 10;            if (result > INT_MAX || result < INT_MIN)                return 0;            x /= 10;        }        return result;    }};

Version 2 String

class Solution {public:    int reverse(int x) {        string s0 = to_string(abs(x));        string s1(s0.rbegin(), s0.rend());        long result = stol(s1);        if (x < 0) result = -result;        return (result > INT_MAX || result < INT_MIN)? 0 : result;    }};

Appendix

• Link: https://leetcode.com/problems/reverse-integer/
• Run Time:
  
  • Version 1: 15ms
  • Version 2: 19ms