在不使用任何额外空格的情况下检查数字是否为回文 |设置 2

给定一个数字“n”，我们的目标是在不使用任何额外空间的情况下找出它是否是回文。我们无法制作新的号码副本。

例子：

Input: n = 2332
Output: Yes it is Palindrome.
Explanation:
original number = 2332
reversed number = 2332
Both are same hence the number is palindrome.

Input: n=1111
Output: Yes it is Palindrome.

Input: n = 1234
Output: No not Palindrome.

其他方法：本文的 Set-1 中讨论了其他递归方法和比较数字的方法。在这里，我们正在讨论其他方法。

方法：这种方法取决于3个主要步骤，找到数字中的位数。将数字从中间分成两部分。注意长度为奇数的情况，在这种情况下，我们将不得不使用中间元素两次。检查两个号码中的号码是否相同。请按照以下步骤解决问题：

• 将变量K初始化为数字n 的长度。
• 将变量ans初始化为0。
• 使用变量i迭代范围[0, K/2)并执行以下任务：
  • 将n%10的值放入变量ans中，然后将n除以10。
• 如果K%2等于1 ，则将n%10的值放入变量ans 中。
• 执行上述步骤后，如果ans等于n那么它是一个回文，否则不是。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find if the number is a
// pallindrome or not
bool isPalindrome(int n)
{
    if (n < 0)
        return false;
    if (n < 10)
        return true;
 
    // Find the length of the number
    int K = ceil(log(n) / log(10));
 
    int ans = 0;
 
    // Partition the number into 2 halves
    for (int i = 0; i < K / 2; i++) {
        ans = ans * 10 + n % 10;
        n = n / 10;
    }
    if (K % 2 == 1)
        ans = ans * 10 + n % 10;
 
    // Equality Condition
    return (ans == n);
}
 
// Driver Code
int main()
{
    isPalindrome(1001) ? cout << "Yes, it is Palindrome"
                       : cout << "No, not Palindrome";
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find if the number is a
// pallindrome or not
static boolean isPalindrome(int n)
{
    if (n < 0)
        return false;
    if (n < 10)
        return true;
 
    // Find the length of the number
    int K = (int) Math.ceil(Math.log(n) / Math.log(10));
 
    int ans = 0;
 
    // Partition the number into 2 halves
    for (int i = 0; i < K / 2; i++) {
        ans = ans * 10 + n % 10;
        n = n / 10;
    }
    if (K % 2 == 1)
        ans = ans * 10 + n % 10;
 
    // Equality Condition
    return (ans == n);
}
 
// Driver Code
public static void main(String[] args)
{
    System.out.print(isPalindrome(1001) ? "Yes, it is Palindrome"
                       : "No, not Palindrome");
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python code for the above approach
import math as Math
 
# Function to find if the number is a
# pallindrome or not
def isPalindrome(n):
    if (n < 0):
        return False
    if (n < 10):
        return True
 
    # Find the length of the number
    K = Math.ceil(Math.log(n) / Math.log(10))
 
    ans = 0
 
    # Partition the number into 2 halves
    for i in range(0, K // 2):
        ans = ans * 10 + n % 10
        n = Math.floor(n / 10)
    if (K % 2 == 1):
        ans = ans * 10 + n % 10
 
    # Equality Condition
    return (ans == n)
 
# Driver Code
print("Yes, it is Palindrome") if isPalindrome(
    1001) else print("No, not Palindrome")
 
# This code is contributed by Saurabh jaiswal

C#

// C# program for the above approach
using System;
 
class GFG
{
 
  // Function to find if the number is a
  // pallindrome or not
  static bool isPalindrome(int n)
  {
    if (n < 0)
      return false;
    if (n < 10)
      return true;
 
    // Find the length of the number
    int K = (int)Math.Ceiling(Math.Log(n) / Math.Log(10));
 
    int ans = 0;
 
    // Partition the number into 2 halves
    for (int i = 0; i < K / 2; i++)
    {
      ans = ans * 10 + n % 10;
      n = n / 10;
    }
    if (K % 2 == 1)
      ans = ans * 10 + n % 10;
 
    // Equality Condition
    return (ans == n);
  }
 
  // Driver Code
  public static void Main()
  {
    Console.Write(isPalindrome(1001) ? "Yes, it is Palindrome" : "No, not Palindrome");
  }
}
 
// This code is contributed by Saurabh Jaiswal

Javascript

       // JavaScript code for the above approach
 
       // Function to find if the number is a
       // pallindrome or not
       function isPalindrome(n) {
           if (n < 0)
               return false;
           if (n < 10)
               return true;
 
           // Find the length of the number
           let K = Math.ceil(Math.log(n) / Math.log(10));
 
           let ans = 0;
 
           // Partition the number into 2 halves
           for (let i = 0; i < K / 2; i++) {
               ans = ans * 10 + n % 10;
               n = Math.floor(n / 10);
           }
           if (K % 2 == 1)
               ans = ans * 10 + n % 10;
 
           // Equality Condition
           return (ans == n);
       }
 
       // Driver Code
       isPalindrome(1001) ? document.write("Yes, it is Palindrome")
           : document.write("No, not Palindrome");
 
 // This code is contributed by Potta Lokesh
   

输出

Yes, it is Palindrome

时间复杂度： O(K)，其中K 是位数
辅助方法： O(1)