数组中最大的回文数
给定一个非负整数数组 arr[] 。任务是找出数组中最大的一个数，也就是回文。如果没有这样的号码，则打印 -1 。
示例:
输入: arr[] = {1，232，54545，999991 }； T3】产量: 54545
输入: arr[] = {1，2，3，4，5，50 }； T3】输出: 5
方法 1:
- 按升序排序数组。
- 从末尾开始遍历数组。
- 第一个数字是回文，这是必须的答案。
- 如果没有找到回文编号，则打印 -1
下面是上述方法的实现:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if n is palindrome
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 digits are
// not same then return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digits from the number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Function to find the largest palindromic number
int largestPalindrome(int A[], int n)
{
// Sort the array
sort(A, A + n);
for (int i = n - 1; i >= 0; --i) {
// If number is palindrome
if (isPalindrome(A[i]))
return A[i];
}
// If no palindromic number found
return -1;
}
// Driver program
int main()
{
int A[] = { 1, 232, 54545, 999991 };
int n = sizeof(A) / sizeof(A[0]);
// print required answer
cout << largestPalindrome(A, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of above approach
import java.util.*;
class GFG
{
// Function to check if n is palindrome
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 digits are
// not same then return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digits from the number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Function to find the largest palindromic number
static int largestPalindrome(int []A, int n)
{
// Sort the array
Arrays.sort(A);
for (int i = n - 1; i >= 0; --i) {
// If number is palindrome
if (isPalindrome(A[i]))
return A[i];
}
// If no palindromic number found
return -1;
}
// Driver program
public static void main(String []args)
{
int []A = { 1, 232, 54545, 999991 };
int n = A.length;
// print required answer
System.out.println(largestPalindrome(A, n));
}
}
// This code is contributed
// by ihritik
Python 3
# Python3 implementation of above approach
# Function to check if n is palindrome
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 digits are
# not same then return false
if (leading != trailing) :
return False
# Removing the leading and trailing
# digits from the number
n = (n % divisor) // 10
# Reducing divisor by a factor
# of 2 as 2 digits are dropped
divisor = divisor // 100
return True
# Function to find the largest
# palindromic number
def largestPalindrome(A, n) :
# Sort the array
A.sort()
for i in range(n - 1, -1, -1) :
# If number is palindrome
if (isPalindrome(A[i])) :
return A[i]
# If no palindromic number found
return -1
# Driver Code
if __name__ == "__main__" :
A = [ 1, 232, 54545, 999991 ]
n = len(A)
# print required answer
print(largestPalindrome(A, n))
# This code is contributed by Ryuga
C
// C# implementation of above approach
using System;
class GFG
{
// Function to check if n is palindrome
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 digits are
// not same then return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digits from the number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Function to find the largest palindromic number
static int largestPalindrome(int []A, int n)
{
// Sort the array
Array.Sort(A);
for (int i = n - 1; i >= 0; --i) {
// If number is palindrome
if (isPalindrome(A[i]))
return A[i];
}
// If no palindromic number found
return -1;
}
// Driver program
public static void Main()
{
int []A = { 1, 232, 54545, 999991 };
int n = A.Length;
// print required answer
Console.WriteLine(largestPalindrome(A, n));
}
}
// This code is contributed
// by ihritik
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of above approach
// Function to check if n is palindrome
function isPalindrome($n)
{
// Find the appropriate divisor
// to extract the leading digit
$divisor = 1;
while ((int)($n / $divisor) >= 10)
$divisor *= 10;
while ($n != 0)
{
$leading = (int)($n / $divisor);
$trailing = $n % 10;
// If first and last digits are
// not same then return false
if ($leading != $trailing)
return false;
// Removing the leading and trailing
// digits from the number
$n = (int)(($n % $divisor) / 10);
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
$divisor = (int)($divisor / 100);
}
return true;
}
// Function to find the largest
// palindromic number
function largestPalindrome($A, $n)
{
// Sort the array
sort($A);
for ($i = $n - 1; $i >= 0; --$i)
{
// If number is palindrome
if (isPalindrome($A[$i]))
return $A[$i];
}
// If no palindromic number found
return -1;
}
// Driver Code
$A = array(1, 232, 54545, 999991);
$n = sizeof($A);
// print required answer
echo largestPalindrome($A, $n);
// This code is contributed
// by Akanksha Rai
?>
java 描述语言
<script>
//Javascript implementation of above approach
//function for sorting
function ssort(a, n) {
var i, j, min, temp;
for (i = 0; i < n - 1; i++) {
min = i;
for (j = i + 1; j < n; j++)
if (a[j] < a[min])
min = j;
temp = a[i];
a[i] = a[min];
a[min] = temp;
}
}
// Function to check if n is palindrome
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 digits are
// not same then return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digits from the number
n = parseInt((n % divisor) / 10);
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = parseInt(divisor / 100);
}
return true;
}
// Function to find the largest palindromic number
function largestPalindrome( A, n)
{
// Sort the array
ssort(A,A.length);
for (var i = n - 1; i >= 0; --i) {
// If number is palindrome
if (isPalindrome(A[i]))
return A[i];
}
// If no palindromic number found
return -1;
}
var A = [ 1, 232, 54545, 999991 ];
var n = A.length;
// print required answer
document.write(largestPalindrome(A, n));
//This code is contributed by SoumikMondal
</script>
Output:
54545
方法二:
- 设置一个变量 currentMax = -1 ，开始遍历数组。
- 如果当前元素arr【I】>currentMax和arr【I】是回文。
- 然后设置 currentMax = arr[i] 。
- 最后打印 currentMax 。
下面是上述方法的实现:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if n is palindrome
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 digits are
// not same then return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digits from the number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Function to find the largest palindromic number
int largestPalindrome(int A[], int n)
{
int currentMax = -1;
for (int i = 0; i < n; i++) {
// If a palindrome larger than the currentMax is found
if (A[i] > currentMax && isPalindrome(A[i]))
currentMax = A[i];
}
// Return the largest palindromic number from the array
return currentMax;
}
// Driver program
int main()
{
int A[] = { 1, 232, 54545, 999991 };
int n = sizeof(A) / sizeof(A[0]);
// print required answer
cout << largestPalindrome(A, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of above approach
import java.util.*;
class GFG
{
// Function to check if n is palindrome
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 digits are
// not same then return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digits from the number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Function to find the largest palindromic number
static int largestPalindrome(int []A, int n)
{
int currentMax = -1;
for (int i = 0; i < n; i++) {
// If a palindrome larger than the currentMax is found
if (A[i] > currentMax && isPalindrome(A[i]))
currentMax = A[i];
}
// Return the largest palindromic number from the array
return currentMax;
}
// Driver program
public static void main(String []args)
{
int []A = { 1, 232, 54545, 999991 };
int n = A.length;
// print required answer
System.out.println(largestPalindrome(A, n));
}
}
// This code is contributed
// by ihritik
Python 3
# Python 3 implementation of above approach
# Function to check if n is palindrome
def isPalindrome(n):
# Find the appropriate divisor
# to extract the leading digit
divisor = 1
while (int(n / divisor) >= 10):
divisor *= 10
while (n != 0):
leading = int(n / divisor)
trailing = n % 10
# If first and last digits are
# not same then return false
if (leading != trailing):
return False
# Removing the leading and trailing
# digits from the number
n = int((n % divisor) / 10)
# Reducing divisor by a factor
# of 2 as 2 digits are dropped
divisor = int(divisor / 100)
return True
# Function to find the largest
# palindromic number
def largestPalindrome(A, n):
currentMax = -1
for i in range(0, n, 1):
# If a palindrome larger than
# the currentMax is found
if (A[i] > currentMax and isPalindrome(A[i])):
currentMax = A[i]
# Return the largest palindromic
# number from the array
return currentMax
# Driver Code
if __name__ == '__main__':
A = [1, 232, 54545, 999991]
n = len(A)
# print required answer
print(largestPalindrome(A, n))
# This code is contributed by
# Surendra_Gangwar
C
// C# implementation of above approach
using System;
class GFG
{
// Function to check if n is palindrome
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 digits are
// not same then return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digits from the number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Function to find the largest palindromic number
static int largestPalindrome(int []A, int n)
{
int currentMax = -1;
for (int i = 0; i < n; i++) {
// If a palindrome larger than the currentMax is found
if (A[i] > currentMax && isPalindrome(A[i]))
currentMax = A[i];
}
// Return the largest palindromic number from the array
return currentMax;
}
// Driver program
public static void Main()
{
int []A = { 1, 232, 54545, 999991 };
int n = A.Length;
// print required answer
Console.WriteLine(largestPalindrome(A, n));
}
}
// This code is contributed
// by ihritik
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of above approach
// Function to check if n is palindrome
function isPalindrome($n)
{
// Find the appropriate divisor
// to extract the leading digit
$divisor = 1;
while ((int)($n / $divisor) >= 10)
$divisor *= 10;
while ($n != 0)
{
$leading = (int)($n / $divisor);
$trailing = $n % 10;
// If first and last digits are
// not same then return false
if ($leading != $trailing)
return false;
// Removing the leading and trailing
// digits from the number
$n = ($n % $divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
$divisor = $divisor / 100;
}
return true;
}
// Function to find the largest
// palindromic number
function largestPalindrome($A, $n)
{
$currentMax = -1;
for ($i = 0; $i < $n; $i++)
{
// If a palindrome larger than
// the currentMax is found
if ($A[$i] > $currentMax &&
isPalindrome($A[$i]))
$currentMax = $A[$i];
}
// Return the largest palindromic
// number from the array
return $currentMax;
}
// Driver Code
$A = array(1, 232, 54545, 999991);
$n = sizeof($A);
// print required answer
echo(largestPalindrome($A, $n));
// This code is contributed
// by Mukul Singh
?>
java 描述语言
<script>
// Javascript implementation of above approach
// Function to check if n is palindrome
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 digits are
// not same then return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digits from the number
n = parseInt((n % divisor) / 10);
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = parseInt(divisor / 100);
}
return true;
}
// Function to find the largest palindromic number
function largestPalindrome(A, n)
{
var currentMax = -1;
for(var i = 0; i < n; i++)
{
// If a palindrome larger than the
// currentMax is found
if (A[i] > currentMax && isPalindrome(A[i]))
currentMax = A[i];
}
// Return the largest palindromic
// number from the array
return currentMax;
}
// Driver code
var A = [ 1, 232, 54545, 999991 ];
var n = A.length;
// Print required answer
document.write( largestPalindrome(A, n));
// This code is contributed by rrrtnx
</script>
Output:
54545
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