N位最小和最大回文

给定数字N。任务是找到N位数可能出现的最小和最大回文数。
例子：

Input: N = 4 
Output: 
Smallest Palindrome = 1001
Largest Palindrome = 9999

Input: N = 5
Output: 
Smallest Palindrome = 10001
Largest Palindrome = 99999

最小的N位回文数：仔细观察，您会发现对于N = 1，最小的回文数将为0。对于其他任何N值，最小的回文数的第一位和最后一位均为1。之间的数字为0。

• 情况1：如果N = 1，则答案将为0。
• 情况2：如果N！= 1，则答案将是(10 ^(N-1) )+ 1。

最大的N位回文数：与上述方法类似，您可以看到，通过在N上附加9，可以获得具有N位的最大回文数。因此，最大的N位回文数将为10 ^N – 1 。
下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to print the smallest and largest
// palindrome with N digits
void printPalindrome(int n)
{
    if (n == 1)
    {
        cout<<"Smallest Palindrome: 0"<

---

Java

// Java implementation of the above approach
class GfG {
 
    // Function to print the smallest and largest
    // palindrome with N digits
    static void printPalindrome(int n)
    {
        if (n == 1)
        {
            System.out.println("Smallest Palindrome: 0");
            System.out.println("Largest Palindrome: 9");
        }
        else
        {
            System.out.println("Smallest Palindrome: "
                    + (int)(Math.pow(10, n - 1)) + 1);
                     
            System.out.println("Largest Palindrome: "
                    + ((int)(Math.pow(10,n)) - 1));
        }
    }
     
    // Driver Code
    public static void main(String[] args) {
        int n = 4;
        printPalindrome(n);
    }
}

Python3

# Python 3 implementation of the above approach
 
from math import pow
 
# Function to print the smallest and largest
# palindrome with N digits
def printPalindrome(n):
    if (n == 1):
        print("Smallest Palindrome: 0")
        print("Largest Palindrome: 9")
    else:
        print("Smallest Palindrome:", int(pow(10, n - 1))+1)
        print("Largest Palindrome:", int(pow(10,n))-1)
     
 
# Driver Code
if __name__ == '__main__':
    n = 4
    printPalindrome(n)
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
 
class GfG
{
 
    // Function to print the smallest and largest
    // palindrome with N digits
    static void printPalindrome(int n)
    {
        if (n == 1)
        {
            Console.WriteLine("Smallest Palindrome: 0");
            Console.WriteLine("Largest Palindrome: 9");
        }
        else
        {
            Console.WriteLine("Smallest Palindrome: "
                    + (int)(Math.Pow(10, n - 1)) + 1);
                     
            Console.WriteLine("Largest Palindrome: "
                    + ((int)(Math.Pow(10,n)) - 1));
        }
    }
     
    // Driver Code
    public static void Main(String[] args)
    {
        int n = 4;
        printPalindrome(n);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

    // Javascript implementation of the above approach
 
    // Function to print the smallest and largest
    // palindrome with N digits
    function printPalindrome(n)
    {
      if (n == 1)
      {
        document.write("Smallest Palindrome: 0<br>");
        document.write("Largest Palindrome: 9");
      }
      else
      {
        document.write("Smallest Palindrome: " + (parseInt(Math.pow(10, n - 1)) + 1));
        document.write("<br>Largest Palindrome: " + parseInt(Math.pow(10, n) - 1));
      }
    }
 
    // Driver Code
    var n = 4;
    printPalindrome(n);
 
// This code is contributed by rrrtnx.
  

输出：

Smallest Palindrome: 1001
Largest Palindrome: 9999

时间复杂度： O(1)