第 N 个偶数长度回文

给定一个数字 n 作为字符串，找到第 n 个偶数长度的正回文数。

例子：

Input : n = "1"
Output : 11
1st even-length palindrome is 11 .

Input : n = "10"
Output : 1001
The first 10 even-length palindrome numbers are 11, 22, 
33, 44, 55, 66, 77, 88, 99 and 1001.

因为，它是一个偶数长度的回文，所以它的前半部分应该等于后半部分的反转，长度将是 2、4、6、8 ...。要评估第 n 个回文数，我们只看第 10 个偶数长度的回文数 11、22、33、44、55、66、77、88、99 和 1001。这里，第 n 个回文数是 nn'，其中 n' 是 n 的倒数。因此我们只需要以连续的方式写 n 和 n' ，其中 n' 是 n 的倒数。

下面是这种方法的实现。

C++

// C++ program to find n=th even length string.
#include 
using namespace std;
 
// Function to find nth even length Palindrome
string evenlength(string n)
{
    // string r to store resultant
    // palindrome. Initialize same as s
    string res = n;
 
    // In this loop string r stores
    // reverse of string s after the
    // string s in consecutive manner .
    for (int j = n.length() - 1; j >= 0; --j)
        res += n[j];
 
    return res;
}
 
// Driver code
int main()
{
    string n = "10";
   
    // Function call
    cout << evenlength(n);
    return 0;
}

Java

// Java program to find nth even length Palindrome
import java.io.*;
 
class GFG
{
    // Function to find nth even length Palindrome
    static String evenlength(String n)
    {
        // string r to store resultant
        // palindrome. Initialize same as s
        String res = n;
 
        // In this loop string r stores
        // reverse of string s after the
        // string s in consecutive manner
        for (int j = n.length() - 1; j >= 0; --j)
            res += n.charAt(j);
 
        return res;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String n = "10";
       
        // Function call
        System.out.println(evenlength(n));
    }
}
 
// Contributed by Pramod Kumar

Python3

# Python3 program to find n=th even
# length string.
import math as mt
 
# Function to find nth even length
# Palindrome
 
 
def evenlength(n):
 
    # string r to store resultant
    # palindrome. Initialize same as s
    res = n
 
    # In this loop string r stores
    # reverse of string s after the
    # string s in consecutive manner .
    for j in range(len(n) - 1, -1, -1):
        res += n[j]
 
    return res
 
 
# Driver code
n = "10"
 
# Function call
print(evenlength(n))
 
# This code is contributed by
# Mohit kumar 29

C#

// C# program to find nth even
// length Palindrome
using System;
 
class GFG {
 
    // Function to find nth even
    // length Palindrome
    static string evenlength(string n)
    {
 
        // string r to store resultant
        // palindrome. Initialize same
        // as s
        string res = n;
 
        // In this loop string r stores
        // reverse of string s after
        // the string s in consecutive
        // manner
        for (int j = n.Length - 1; j >= 0; --j)
            res += n[j];
 
        return res;
    }
 
    // Driver code
    public static void Main()
    {
        string n = "10";
 
        // Function call
        Console.WriteLine(evenlength(n));
    }
}
 
// This code is contributed by vt_m.

PHP

= 0; --$j)
        $res = $res . $n[$j];
 
    return $res;
}
 
// Driver code
$n = "10";
 
// Function call
echo evenlength($n);
 
// This code is contributed by ita_c
?>

Javascript

// Javascript program to find nth even length Palindrome
     
    // Function to find nth even length Palindrome
    function evenlength(n)
    {
        // string r to store resultant
        // palindrome. Initialize same as s
        let res = n;
  
        // In this loop string r stores
        // reverse of string s after the
        // string s in consecutive manner
        for (let j = n.length - 1; j >= 0; --j)
            res += n[j];
  
        return res;
    }
     
    // Driver code
    let n = "10";
    // Function call
    document.write(evenlength(n));
     
    //This code is contributed by avanitrachhadiya2155
     

输出

1001

时间复杂度： O(n)