不使用任何多余的空格来检查一个数字是不是回文
原文:https://www . geeksforgeeks . org/check-number-回文-非不使用-extra-space/
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。我们不能复制一份新的号码。 例:
Input : 2332
Output : Yes it is Palindrome.
Explanation:
original number = 2332
reversed number = 2332
Both are same hence the number is palindrome.
Input :1111
Output :Yes it is Palindrome.
Input : 1234
Output : No not Palindrome.
一个递归的解决方案将在下面的文章中讨论。 检查一个数字是否是回文 在这篇文章中讨论了一个不同的解决方案。 1)我们可以比较第一个数字和最后一个数字，然后我们重复这个过程。 2)对于第一个数字，我们需要数字的顺序。比方说，12321。把这个除以 10000 会得到领先的 1。尾部的 1 可以通过取 10 的 mod 来检索。 3)现在，把这个减少到 232。
(12321 % 10000)/10 = (2321)/10 = 232
4)现在，10000 需要减少 100 倍。 以下是上述算法的实现:
C++
// C++ program to find number is palindrome
// or not without using any extra space
#include <bits/stdc++.h>
using namespace std;
bool isPalindrome(int);
bool isPalindrome(int n)
{
// Find the appropriate divisor
// to extract the leading digit
int divisor = 1;
while (n / divisor >= 10)
divisor *= 10;
while (n != 0)
{
int leading = n / divisor;
int trailing = n % 10;
// If first and last digit
// not same return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digit from number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Driver code
int main()
{
isPalindrome(1001) ? cout << "Yes, it is Palindrome" :
cout << "No, not Palindrome";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find number is palindrome
// or not without using any extra space
public class GFG
{
static boolean isPalindrome(int n)
{
// Find the appropriate divisor
// to extract the leading digit
int divisor = 1;
while (n / divisor >= 10)
divisor *= 10;
while (n != 0)
{
int leading = n / divisor;
int trailing = n % 10;
// If first and last digit
// not same return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digit from number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Driver code
public static void main(String args[])
{
if(isPalindrome(1001))
System.out.println("Yes, it is Palindrome");
else
System.out.println("No, not Palindrome");
}
}
// This code is contributed by Sumit Ghosh
Python 3
# Python program to find number
# is palindrome or not without
# using any extra space
# Function to check if given number
# is palindrome or not without
# using the extra space
def isPalindrome(n):
# Find the appropriate divisor
# to extract the leading digit
divisor = 1
while (n / divisor >= 10):
divisor *= 10
while (n != 0):
leading = n // divisor
trailing = n % 10
# If first and last digit
# not same return false
if (leading != trailing):
return False
# Removing the leading and
# trailing digit from number
n = (n % divisor)//10
# Reducing divisor by a factor
# of 2 as 2 digits are dropped
divisor = divisor/100
return True
# Driver code
if(isPalindrome(1001)):
print('Yes, it is palindrome')
else:
print('No, not palindrome')
# This code is contributed by Danish Raza
C
// C# program to find number
// is palindrome or not without
// using any extra space
using System;
class GFG
{
static bool isPalindrome(int n)
{
// Find the appropriate
// divisor to extract
// the leading digit
int divisor = 1;
while (n / divisor >= 10)
divisor *= 10;
while (n != 0)
{
int leading = n / divisor;
int trailing = n % 10;
// If first and last digit
// not same return false
if (leading != trailing)
return false;
// Removing the leading and
// trailing digit from number
n = (n % divisor) / 10;
// Reducing divisor by
// a factor of 2 as 2
// digits are dropped
divisor = divisor / 100;
}
return true;
}
// Driver code
static public void Main ()
{
if(isPalindrome(1001))
Console.WriteLine("Yes, it " +
"is Palindrome");
else
Console.WriteLine("No, not " +
"Palindrome");
}
}
// This code is contributed by m_kit
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find
// number is palindrome
// or not without using
// any extra space
function isPalindrome($n)
{
// Find the appropriate divisor
// to extract the leading digit
$divisor = 1;
while ($n / $divisor >= 10)
$divisor *= 10;
while ($n != 0)
{
$leading = floor($n / $divisor);
$trailing = $n % 10;
// If first and last digit
// not same return false
if ($leading != $trailing)
return false;
// Removing the leading and
// trailing digit from number
$n = ($n % $divisor) / 10;
// Reducing divisor by a
// factor of 2 as 2 digits
// are dropped
$divisor = $divisor / 100;
}
return true;
}
// Driver code
if(isPalindrome(1001) == true)
echo "Yes, it is Palindrome" ;
else
echo "No, not Palindrome";
// This code is contributed by ajit
?>
java 描述语言
<script>
// javascript program to find number is palindrome
// or not without using any extra space
function isPalindrome(n) {
// Find the appropriate divisor
// to extract the leading digit
var divisor = 1;
while (parseInt(n / divisor) >= 10)
divisor *= 10;
while (n != 0) {
var leading = parseInt(n / divisor);
var trailing = n % 10;
// If first and last digit
// not same return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digit from number
n = parseInt((n % divisor) / 10);
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Driver code
if (isPalindrome(1001))
document.write("Yes, it is Palindrome");
else
document.write("No, not Palindrome");
// This code is contributed by todaysgaurav
</script>
输出:
Yes, it is Palindrome
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