求第 k 个最小奇数长度回文数
原文:https://www . geesforgeks . org/find-the-kth-minist-odd-length-回文-number/
给定一个正整数 K ，任务是找到 K 第个最小的回文号个奇数长度。
示例:
输入: K = 5 输出: 5 解释: 奇数长度的回文数为{1，2，3，4，5，6，7，…，}。第 5 个最小回文数是 5。
输入:K = 10 T3】输出: 101
方法:给定的问题可以基于以下观察来解决:
- 长度为 1 的第一个回文数字是 1、2、3、4、5、6、7、8 和 9。
- 长度为 3 的第一个回文数是 101，这是第十个最小的奇数长度回文数。同样，第 11 次、第 12 次、第 13 次、…、第 99 次最小回文号分别为 111、121、131 …、999。
- 因此，除最后一位数字外，第 K 个个最小奇数长度回文数可以通过连接 K 和 K 的倒数构成。
从上面的观察来看， Kth 最小的奇数长度回文号是通过追加除了最后一个在 K 末尾的数字之外的 K 的所有数字的倒数给出的。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the Kth smallest
// odd length palindrome
int oddLengthPalindrome(int k)
{
// Store the original number K
int palin = k;
// Removing the last digit of K
k = k / 10;
// Generate the palindrome by
// appending the reverse of K
// except last digit to itself
while (k > 0)
{
// Find the remainder
int rev = k % 10;
// Add the digit to palin
palin = (palin * 10) + rev;
// Divide K by 10
k = k / 10;
}
// Return the resultant palindromic
// number formed
return palin;
}
// Driver Code
int main()
{
int k = 504;
cout << oddLengthPalindrome(k);
}
// This code is contributed by rishavmahato348
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
import java.lang.*;
class GFG{
// Function to find the Kth smallest
// odd length palindrome
static int oddLengthPalindrome(int k)
{
// Store the original number K
int palin = k;
// Removing the last digit of K
k = k / 10;
// Generate the palindrome by
// appending the reverse of K
// except last digit to itself
while (k > 0)
{
// Find the remainder
int rev = k % 10;
// Add the digit to palin
palin = (palin * 10) + rev;
// Divide K by 10
k = k / 10;
}
// Return the resultant palindromic
// number formed
return palin;
}
// Driver Code
public static void main(String[] args)
{
int k = 504;
System.out.println(oddLengthPalindrome(k));
}
}
// This code is contributed by Sudhanshu Bhagat & Govind Choudhary
Python 3
# Python3 program for the above approach
# Function to find the Kth smallest
# odd length palindrome number
def oddLengthPalindrome(K):
# Store the original number K
palin = K
# Removing the last digit of K
K = K // 10
# Generate the palindrome by
# appending the reverse of K
# except last digit to itself
while (K > 0):
# Find the remainder
rev = K % 10
# Add the digit to palin
palin = palin * 10 + rev
# Divide K by 10
K = K // 10
# Return the resultant palindromic
# number formed
return palin
# Driver Code
if __name__ == '__main__':
K = 504
print(oddLengthPalindrome(K))
#Contributed by Govind Choudhary & Pallav Pushparaj
C
// C# program for the above approach
using System;
class GFG{
// Function to find the Kth smallest
// palindrome of odd length
static int oddLengthPalindrome(int k)
{
// Store the original number K
int palin = k;
// Removing the last digit of K
k = k / 10;
// Generate the palindrome by
// appending the reverse of K
// except last digit to itself
while (k > 0)
{
// Find the remainder
int rev = k % 10;
// Add the digit to palin
palin = (palin * 10) + rev;
// Divide K by 10
k = k / 10;
}
// Return the resultant palindromic
// number formed
return palin;
}
// Driver Code
static void Main(string[] args)
{
int k = 504;
Console.WriteLine(oddLengthPalindrome(k));
}
}
// This code is contributed by Sudhanshu Bhagat & Govind Choudhary
java 描述语言
<script>
// JavaScript program for the above approach
// Function to find the Kth smallest
// odd length palindrome
function oddLengthPalindrome(k)
{
// Store the original number K
let palin = k;
// Removing the last digit of K
k = Math.floor(k / 10);
// Generate the palindrome by
// appending the reverse of K
// except last digit to itself
while (k > 0)
{
// Find the remainder
let rev = k % 10;
// Add the digit to palin
palin = (palin * 10) + rev;
// Divide K by 10
k = Math.floor(k / 10);
}
// Return the resultant palindromic
// number formed
return palin;
}
// Driver Code
let k = 504;
document.write(oddLengthPalindrome(k));
// This code is contributed by sanjoy_62.
</script>
Output:
50405
时间复杂度:O(log10K) 辅助空间: O(1)
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