Reverse Integer LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Reverse Integer in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Reverse Integer– LeetCode Problem

Reverse Integer– LeetCode Problem

Problem:

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 - 1], then return 0.

Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

Example 1:

Input: x = 123
Output: 321

Example 2:

Input: x = -123
Output: -321

Example 3:

Input: x = 120
Output: 21

Constraints:

• -231 <= x <= 231 - 1

Reverse Integer– LeetCode Solutions

class Solution {
 public:
  int reverse(int x) {
    long ans = 0;

    while (x) {
      ans = ans * 10 + x % 10;
      x /= 10;
    }

    return (ans < INT_MIN || ans > INT_MAX) ? 0 : ans;
  }
};

class Solution {
  public int reverse(int x) {
    long ans = 0;

    while (x != 0) {
      ans = ans * 10 + x % 10;
      x /= 10;
    }

    return (ans < Integer.MIN_VALUE || ans > Integer.MAX_VALUE) ? 0 : (int) ans;
  }
}

class Solution:
  def reverse(self, x: int) -> int:
    ans = 0
    sign = -1 if x < 0 else 1
    x *= sign

    while x:
      ans = ans * 10 + x % 10
      x //= 10

    return 0 if ans < -2**31 or ans > 2**31 - 1 else sign * ans