第 n 个偶数长度回文
原文:https://www.geeksforgeeks.org/nth-even-length-palindrome/
给定一个 n 为字符串的数，求第 n 个偶数长度的正回文数。
示例:
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倒数，长度是 2，4，6，8 …为了计算第 n 个回文，让我们看看第 10 个偶数长度的回文数字 11，22，33，44，55，66，77，88，99 和 1001。这里，第 n 个回文是 nn '其中 n '是 n 的反义词。因此我们只需要以连续的方式写 n 和 n ’,其中 n’是 n 的反义词。
下面是这个方法的实现。
C++
// C++ program to find n=th even length string.
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find n=th even length string.
// Function to find nth even length Palindrome
function 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 = strlen($n) - 1; $j >= 0; --$j)
$res = $res . $n[$j];
return $res;
}
// Driver code
$n = "10";
// Function call
echo evenlength($n);
// This code is contributed by ita_c
?>
java 描述语言
<script>
// 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
</script>
Output
1001
时间复杂度: O(n)
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