Algorithm

Reverse Integer - https://leetcode.com/problems/reverse-integer/
 
Medium

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

 

Code Examples

#1 Code Example with Javascript Programming

Code - Javascript Programming


var reverse = function(x) {
    let rev = 0;
    let rem;
    let isNegative = x  <  0;
    
    if (isNegative)
        x = (-1) * x;
 
    while(x != 0 ) {
      rem = x % 10;
      x = parseInt(x / 10); 
      rev = parseInt(rev + '' + rem);
    }
    
    if (rev > 0x7FFFFFFF)
        return 0;
    
    if (isNegative)
        return (-1) * rev;
    
    return rev;
};
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Input

x
–
+
cmd
123

Output

x
–
+
cmd
321

#2 Code Example with C Programming

Code - C Programming


int reverse(int x) {
    int d, k = 0;
    while (x) {
        d = x % 10;
        if ((x > 0 && k > (0x7fffffff - d) / 10) ||
            (x  <  0 && k < ((signed)0x80000000 - d) / 10)) {
            return 0;
        }
        k = k * 10 + d;
        x = x / 10;
    }
    return k;
}
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Input

x
–
+
cmd
-123

Output

x
–
+
cmd
-321

#3 Code Example with C++ Programming

Code - C++ Programming


class Solution {
public:
    int reverse(int x) {
        string s = to_string(x);
        int base = 1;
        int i = s[0] == '-' ? 1 : 0;
        int res = 0;
        while(i  <  s.size()){
            if(base == 1000000000 && (s[i] - '0' > 2 || INT_MAX - (s[i] - '0') * base < res)) return 0;
            res += (s[i] - '0'> * base;
            base *= 10;
            i++;
        }
        return s[0] == '-' ? -res : res;
    }
};
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Input

x
–
+
cmd
120

Output

x
–
+
cmd
21

#4 Code Example with C# Programming

Code - C# Programming


namespace LeetCode
{
    public class _007_ReverseInteger
    {
        public int Reverse(int x)
        {
            var result = 0;

            var positiveOverflow = int.MaxValue / 10;
            var nagativeOverflow = int.MinValue / 10;

            for (; x != 0; x /= 10)
            {
                if (result > positiveOverflow || result  <  nagativeOverflow)
                {
                    return 0;
                }
                result = result * 10 + x % 10;
            }

            return result;
        }
    }
}
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Input

x
–
+
cmd
123

Output

x
–
+
cmd
321

#5 Code Example with Java Programming

Code - Java Programming


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

x
–
+
cmd
-123

Output

x
–
+
cmd
-321

#6 Code Example with Python Programming

Code - Python Programming


class Solution:
    def reverse(self, x):
        x = int(str(x)[::-1]) if x >= 0 else int("-" + str(x)[::-1][:-1]); return -2 ** 31 <= x <= 2 ** 31 - 1 and x or 0
Copy The Code & Try With Live Editor

Input

x
–
+
cmd
120

Output

x
–
+
cmd
21
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Demonstration

Leetcode - Reverse Integer Solution in Javascript

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