给定一个数字，找到比这个数字大的下一个最小回文。例如，如果输入数字为“2 3 5 4 5”，则输出应为“2 3 6 3 2”。如果输入数字为“9 9 9”，则输出应为“1 0 0 1”。 假设输入是一个数组。数组中的每个条目代表输入数字中的一个数字。让数组为’num[]，数组大小为’n’
null
可能有三种不同类型的输入需要分别处理。 1) 输入的数字是回文的，全部为9。例如“9”。输出应为“1 0 1” 2) 输入的数字不是回文。例如“1234”。输出应为“1 3 1” 3) 输入的数字是回文的，并不是所有的9。例如“1 2 1”。输出应为“1 3 1”。
输入类型1的解决方案很简单。输出包含n+1位，其中角位为1，角位之间的所有数字均为0。 现在让我们先谈谈输入类型2和3。如何将给定的数字转换为更大的回文？为了理解解决方案，让我们首先定义以下两个术语：
左侧： 给定数字的左半部分。“123456”的左边是“1233”，而“123445”的左边是“1232”
右侧： 给定数字的右半部分。“123456”的右边是“456”，而“123445”的右边是“45”
要转换为回文，我们可以取其左侧的镜像，也可以取其右侧的镜像。然而，如果我们照右边的镜子，那么这样形成的回文不一定是下一个更大的回文。所以，我们必须把左边的镜子复制到右边。但有些案件必须以不同的方式处理。请参见以下步骤。 我们将从两个指数i和j开始。i指向两个中间元素（或者在n为奇数的情况下指向中间元素周围的两个元素）。我们一个接一个地把i和j分开。
第一步。 首先，忽略左侧与右侧相应部分相同的部分。例如，如果数字是“8 3” 4 2 2 4 6.9〃，我们忽略了中间的四个元素。我现在指向元素3，j现在指向元素6。
第二步。 在步骤1之后，出现以下情况： 案例1： i&j指数跨越国界。 当输入的数字是回文时，就会出现这种情况。在这种情况下，我们只需在中间数字上加1（或在n为偶数的情况下加1），将进位传播到左侧的MSB位，同时将左侧的镜像复制到右侧。 例如，如果给定的数字是“1 2 9 2 1”，我们将9增加到10并传播进位。所以这个数字变成了“1 3 0 3 1”
案例2： 在左侧和右侧之间有不相同的数字。所以，我们只需将左侧镜像到右侧，尽量减少形成的数量，以保证回文的最小值。 在这种情况下，可能会有 两个子案例 .
2.1) 将左侧复制到右侧就足够了，我们不需要增加任何数字，结果只是左侧的镜像。下面是这个子案例的一些例子。 “7”的下一个回文 8. “3 3 2”是“7 8 3 8 7” “12”的下一个回文 5. “3 2 2”是“1 2 5 2 1” “14”的下一个回文 5. 8 7 6 7 8 3 2”是“1 4 5 8 7 7 8 5 4 1” 我们如何检查这个子案例？我们只需要检查第1步中被忽略部分后面的数字。上述示例中突出显示了该数字。如果这个数字大于右边数字中相应的数字，那么将左边复制到右边就足够了，我们不需要做任何其他事情。
2.2) 将左侧复制到右侧是不够的。当上面定义的左侧数字较小时，就会发生这种情况。下面是一些例子。 “7”的下一个回文 1. “3 3 2”是“7 1 4 1 7” “12”的下一个回文 3. “4 6 2 8”是“1 2 3 5 3 2 1” “9 4”的下一个回文 1. 8 7 9 7 8 3 2”是“9 4 1 8 0 8 1 4 9” 我们像案例1一样处理这个子案例。我们只需在中间数字（或n为偶数时的数字）上加1，将进位传播到左侧的MSB位，同时将左侧的镜像复制到右侧。
方法1： 寻找次最小回文数的基本方法。
C++
#include <bits/stdc++.h> using namespace std; // Function to check whether number is palindrome or not int isPalindrome( int num) { // Declaring variables int n, k, rev = 0; // storing num in n so that we can compare it later n = num; // while num is not 0 we find its reverse and store in // rev while (num != 0) { k = num % 10; rev = (rev * 10) + k; num = num / 10; } // check if num and its reverse are same if (n == rev) { return 1; } else { return 0; } } int main() { // Take any number to find its next palindrome number int num = 9687; // If number is not Palindrome we go to the next number // using while loop while (!isPalindrome(num)) { num = num + 1; } // now we get the next Palindrome so let's print it cout << "Next Palindrome :" ; cout << num; return 0; } // Contribute by :- Tejas Bhavsar |
JAVA
import java.io.*; class GFG { // Function to check whether number is palindrome or not static int isPalindrome( int num) { // Declaring variables int n, k, rev = 0 ; // storing num in n so that we can compare it later n = num; // while num is not 0 we find its reverse and store // in rev while (num != 0 ) { k = num % 10 ; rev = (rev * 10 ) + k; num = num / 10 ; } // check if num and its reverse are same if (n == rev) { return 1 ; } else { return 0 ; } } // Driver code public static void main(String[] args) { // Take any number to find its next palindrome // number int num = 9687 ; // If number is not Palindrome we go to the next // number using while loop while (isPalindrome(num) == 0 ) { num = num + 1 ; } // now we get the next Palindrome so let's print it System.out.print( "Next Palindrome :" ); System.out.print(num); } } // This code is contributed by subhammahato348. |
Python3
# Program to print find next palindrome # number greater than given number. # function to check a number is # palindrome or not def isPalindrome(num): # Declaring variables # storing num in n so that we can compare it later n = num rev = 0 # while num is not 0 we find its reverse and store # in rev while (num > 0 ): k = num % 10 rev = (rev * 10 ) + k num = num / / 10 # check if num and its reverse are same if (n = = rev): return True else : return False # input number num = 9687 # start check from next num; num = num + 1 # Loop checks all numbers from given no. # (num + 1) to next palindrome no. while ( True ): if (isPalindrome(num)): break num = num + 1 # printing the next palindrome print ( "Next Palindrome :" ) print (num) # This code is contributed by sidharthsingh7898. |
C#
using System; class GFG { // Function to check whether number is palindrome or not static int isPalindrome( int num) { // Declaring variables int n, k, rev = 0; // storing num in n so that we can compare it later n = num; // while num is not 0 we find its reverse and store // in rev while (num != 0) { k = num % 10; rev = (rev * 10) + k; num = num / 10; } // check if num and its reverse are same if (n == rev) { return 1; } else { return 0; } } // Driver code public static void Main() { // Take any number to find its next palindrome // number int num = 9687; // If number is not Palindrome we go to the next // number using while loop while (isPalindrome(num) == 0) { num = num + 1; } // now we get the next Palindrome so let's print it Console.Write( "Next Palindrome :" ); Console.Write(num); } } // This code is contributed by subhammahato348. |
Javascript
<script> // Javascript program for the above approach // Function to check whether number is palindrome or not function isPalindrome(num) { // Declaring variables let n, k, rev = 0; // storing num in n so that we can compare it later n = num; // while num is not 0 we find its reverse and store // in rev while (num != 0) { k = num % 10; rev = (rev * 10) + k; num = Math.floor(num / 10); } // check if num and its reverse are same if (n == rev) { return 1; } else { return 0; } } // Driver Code // Take any number to find its next palindrome // number let num = 9687; // If number is not Palindrome we go to the next // number using while loop while (isPalindrome(num) == 0) { num = num + 1; } // now we get the next Palindrome so let's print it document.write( "Next Palindrome :" ); document.write(num); // This code is contributed by splevel62. </script> |
输出
Next Palindrome :9779
时间复杂性： O（num*|num |）
C++
#include <iostream> using namespace std; // Utility that prints out an array on a line void printArray( int arr[], int n) { int i; for (i = 0; i < n; i++) printf ( "%d " , arr[i]); printf ( "" ); } // A utility function to check if num has all 9s int AreAll9s( int * num, int n ) { int i; for (i = 0; i < n; ++i) if (num[i] != 9) return 0; return 1; } // Returns next palindrome of a given number num[]. // This function is for input type 2 and 3 void generateNextPalindromeUtil ( int num[], int n ) { // Find the index of mid digit int mid = n / 2; // A bool variable to check if copy of left // side to right is sufficient or not bool leftsmaller = false ; // End of left side is always 'mid -1' int i = mid - 1; // Beginning of right side depends // if n is odd or even int j = (n % 2) ? mid + 1 : mid; // Initially, ignore the middle same digits while (i >= 0 && num[i] == num[j]) i--, j++; // Find if the middle digit(s) need to be // incremented or not (or copying left // side is not sufficient) if (i < 0 || num[i] < num[j]) leftsmaller = true ; // Copy the mirror of left to tight while (i >= 0) { num[j] = num[i]; j++; i--; } // Handle the case where middle digit(s) must // be incremented. This part of code is for // CASE 1 and CASE 2.2 if (leftsmaller == true ) { int carry = 1; i = mid - 1; // If there are odd digits, then increment // the middle digit and store the carry if (n % 2 == 1) { num[mid] += carry; carry = num[mid] / 10; num[mid] %= 10; j = mid + 1; } else j = mid; // Add 1 to the rightmost digit of the // left side, propagate the carry towards // MSB digit and simultaneously copying // mirror of the left side to the right side. while (i >= 0) { num[i] += carry; carry = num[i] / 10; num[i] %= 10; // Copy mirror to right num[j++] = num[i--]; } } } // The function that prints next palindrome // of a given number num[] with n digits. void generateNextPalindrome( int num[], int n) { int i; printf ( "Next palindrome is:" ); // Input type 1: All the digits are 9, simply o/p 1 // followed by n-1 0's followed by 1. if (AreAll9s(num, n)) { printf ( "1 " ); for (i = 1; i < n; i++) printf ( "0 " ); printf ( "1" ); } // Input type 2 and 3 else { generateNextPalindromeUtil(num, n); // print the result printArray (num, n); } } // Driver code int main() { int num[] = { 9, 4, 1, 8, 7, 9, 7, 8, 3, 2, 2 }; int n = sizeof (num) / sizeof (num[0]); generateNextPalindrome(num, n); return 0; } // This code is contributed by rohan07 |
C
#include <stdio.h> / / A utility function to print an array void printArray ( int arr[], int n); / / A utility function to check if num has all 9s int AreAll9s ( int num[], int n ); / / Returns next palindrome of a given number num[]. / / This function is for input type 2 and 3 void generateNextPalindromeUtil ( int num[], int n ) { / / find the index of mid digit int mid = n / 2 ; / / A bool variable to check if copy of left side to right is sufficient or not bool leftsmaller = false; / / end of left side is always 'mid -1' int i = mid - 1 ; / / Beginning of right side depends if n is odd or even int j = (n % 2 )? mid + 1 : mid; / / Initially, ignore the middle same digits while (i > = 0 && num[i] = = num[j]) i - - ,j + + ; / / Find if the middle digit(s) need to be incremented or not ( or copying left / / side is not sufficient) if ( i < 0 || num[i] < num[j]) leftsmaller = true; / / Copy the mirror of left to tight while (i > = 0 ) { num[j] = num[i]; j + + ; i - - ; } / / Handle the case where middle digit(s) must be incremented. / / This part of code is for CASE 1 and CASE 2.2 if (leftsmaller = = true) { int carry = 1 ; i = mid - 1 ; / / If there are odd digits, then increment / / the middle digit and store the carry if (n % 2 = = 1 ) { num[mid] + = carry; carry = num[mid] / 10 ; num[mid] % = 10 ; j = mid + 1 ; } else j = mid; / / Add 1 to the rightmost digit of the left side, propagate the carry / / towards MSB digit and simultaneously copying mirror of the left side / / to the right side. while (i > = 0 ) { num[i] + = carry; carry = num[i] / 10 ; num[i] % = 10 ; num[j + + ] = num[i - - ]; / / copy mirror to right } } } / / The function that prints next palindrome of a given number num[] / / with n digits. void generateNextPalindrome( int num[], int n ) { int i; printf( "Next palindrome is:" ); / / Input type 1 : All the digits are 9 , simply o / p 1 / / followed by n - 1 0 's followed by 1. if ( AreAll9s( num, n ) ) { printf( "1 " ); for ( i = 1 ; i < n; i + + ) printf( "0 " ); printf( "1" ); } / / Input type 2 and 3 else { generateNextPalindromeUtil ( num, n ); / / print the result printArray (num, n); } } / / A utility function to check if num has all 9s int AreAll9s( int * num, int n ) { int i; for ( i = 0 ; i < n; + + i ) if ( num[i] ! = 9 ) return 0 ; return 1 ; } / * Utility that prints out an array on a line * / void printArray( int arr[], int n) { int i; for (i = 0 ; i < n; i + + ) printf( "%d " , arr[i]); printf( "" ); } / / Driver Program to test above function int main() { int num[] = { 9 , 4 , 1 , 8 , 7 , 9 , 7 , 8 , 3 , 2 , 2 }; int n = sizeof (num) / sizeof(num[ 0 ]); generateNextPalindrome( num, n ); return 0 ; } |
JAVA
// Java program to find next smallest // palindrome public class nextplaindrome { // Returns next palindrome of a given // number num[]. This function is for // input type 2 and 3 static void generateNextPalindromeUtil( int num[], int n) { int mid = n / 2 ; // end of left side is always 'mid -1' int i = mid - 1 ; // Beginning of right side depends // if n is odd or even int j = (n % 2 == 0 ) ? mid : mid + 1 ; // A bool variable to check if copy of left // side to right // is sufficient or not boolean leftsmaller = false ; // Initially, ignore the middle same digits while (i >= 0 && num[i] == num[j]) { i--; j++; } // Find if the middle digit(s) need to // be incremented or not (or copying left // side is not sufficient) if (i < 0 || num[i] < num[j]) { leftsmaller = true ; } // Copy the mirror of left to tight while (i >= 0 ) { num[j++] = num[i--]; } // Handle the case where middle digit(s) // must be incremented. This part of code // is for CASE 1 and CASE 2.2 if (leftsmaller) { int carry = 1 ; // If there are odd digits, then increment // the middle digit and store the carry if (n % 2 == 1 ) { num[mid] += 1 ; carry = num[mid] / 10 ; num[mid] %= 10 ; } i = mid - 1 ; j = (n % 2 == 0 ? mid : mid + 1 ); // Add 1 to the rightmost digit of the left // side, propagate the carry towards MSB digit // and simultaneously copying mirror of the // left side to the right side. //when carry is zero no need to loop through till i>=0 while (i >= 0 && carry> 0 ) { num[i] = num[i] + carry; carry = num[i] / 10 ; num[i] %= 10 ; num[j] = num[i]; // copy mirror to right i--; j++; } } } // The function that prints next palindrome // of a given number num[] with n digits. static void generateNextPalindrome( int num[], int n) { System.out.println( "Next Palindrome is:" ); // Input type 1: All the digits are 9, // simply o/p 1 followed by n-1 0's // followed by 1. if (isAll9(num, n)) { System.out.print( "1" ); for ( int i = 0 ; i < n - 1 ; i++) System.out.print( "0" ); System.out.println( "1" ); } // Input type 2 and 3 else { generateNextPalindromeUtil(num, n); printarray(num); } } // A utility function to check if num has all 9s static boolean isAll9( int num[], int n) { for ( int i = 0 ; i < n; i++) if (num[i] != 9 ) return false ; return true ; } /* Utility that prints out an array on a line */ static void printarray( int num[]) { for ( int i = 0 ; i < num.length; i++) System.out.print(num[i]); System.out.println(); } public static void main(String[] args) { int num[] = { 9 , 4 , 1 , 8 , 7 , 9 , 7 , 8 , 3 , 2 , 2 }; generateNextPalindrome(num, num.length); } } |
Python3
# Returns next palindrome of a given number num[]. # This function is for input type 2 and 3 def generateNextPalindromeUtil (num, n) : # find the index of mid digit mid = int (n / 2 ) # A bool variable to check if copy of left # side to right is sufficient or not leftsmaller = False # end of left side is always 'mid -1' i = mid - 1 # Beginning of right side depends # if n is odd or even j = mid + 1 if (n % 2 ) else mid # Initially, ignore the middle same digits while (i > = 0 and num[i] = = num[j]) : i - = 1 j + = 1 # Find if the middle digit(s) need to be # incremented or not (or copying left # side is not sufficient) if ( i < 0 or num[i] < num[j]): leftsmaller = True # Copy the mirror of left to tight while (i > = 0 ) : num[j] = num[i] j + = 1 i - = 1 # Handle the case where middle # digit(s) must be incremented. # This part of code is for CASE 1 and CASE 2.2 if (leftsmaller = = True ) : carry = 1 i = mid - 1 # If there are odd digits, then increment # the middle digit and store the carry if (n % 2 = = 1 ) : num[mid] + = carry carry = int (num[mid] / 10 ) num[mid] % = 10 j = mid + 1 else : j = mid # Add 1 to the rightmost digit of the # left side, propagate the carry # towards MSB digit and simultaneously # copying mirror of the left side # to the right side. while (i > = 0 ) : num[i] + = carry carry = int (num[i] / 10 ) num[i] % = 10 num[j] = num[i] # copy mirror to right j + = 1 i - = 1 # The function that prints next # palindrome of a given number num[] # with n digits. def generateNextPalindrome(num, n ) : print ( "Next palindrome is:" ) # Input type 1: All the digits are 9, simply o/p 1 # followed by n-1 0's followed by 1. if ( AreAll9s( num, n ) = = True ) : print ( "1" ) for i in range ( 1 , n): print ( "0" ) print ( "1" ) # Input type 2 and 3 else : generateNextPalindromeUtil ( num, n ) # print the result printArray (num, n) # A utility function to check if num has all 9s def AreAll9s(num, n ): for i in range ( 1 , n): if ( num[i] ! = 9 ) : return 0 return 1 # Utility that prints out an array on a line def printArray(arr, n): for i in range ( 0 , n): print ( int (arr[i]),end = " " ) print () # Driver Program to test above function if __name__ = = "__main__" : num = [ 9 , 4 , 1 , 8 , 7 , 9 , 7 , 8 , 3 , 2 , 2 ] n = len (num) generateNextPalindrome( num, n ) # This code is contributed by Smitha Dinesh Semwal |
C#
// C# program to find next smallest palindrome using System; public class GFG { // Returns next palindrome of a given // number num[]. This function is for // input type 2 and 3 static void generateNextPalindromeUtil( int []num, int n) { int mid = n / 2; // end of left side is always 'mid -1' int i = mid - 1; // Beginning of right side depends // if n is odd or even int j = (n % 2 == 0) ? mid : mid + 1; // A bool variable to check if copy of left // side to right // is sufficient or not bool leftsmaller = false ; // Initially, ignore the middle same digits while (i >= 0 && num[i] == num[j]) { i--; j++; } // Find if the middle digit(s) need to // be incremented or not (or copying left // side is not sufficient) if (i < 0 || num[i] < num[j]) { leftsmaller = true ; } // Copy the mirror of left to tight while (i >= 0) { num[j++] = num[i--]; } // Handle the case where middle digit(s) // must be incremented. This part of code // is for CASE 1 and CASE 2.2 if (leftsmaller) { int carry = 1; // If there are odd digits, then increment // the middle digit and store the carry if (n % 2 == 1) { num[mid] += 1; carry = num[mid] / 10; num[mid] %= 10; } i = mid - 1; j = (n % 2 == 0 ? mid : mid + 1); // Add 1 to the rightmost digit of the left // side, propagate the carry towards MSB digit // and simultaneously copying mirror of the // left side to the right side. while (i >= 0) { num[i] = num[i] + carry; carry = num[i] / 10; num[i] %= 10; num[j] = num[i]; // copy mirror to right i--; j++; } } } // The function that prints next palindrome // of a given number num[] with n digits. static void generateNextPalindrome( int []num, int n) { Console.WriteLine( "Next Palindrome is:" ); // Input type 1: All the digits are 9, // simply o/p 1 followed by n-1 0's // followed by 1. if (isAll9(num, n)) { Console.Write( "1" ); for ( int i = 0; i < n - 1; i++) Console.Write( "0" ); Console.Write( "1" ); } // Input type 2 and 3 else { generateNextPalindromeUtil(num, n); printarray(num); } } // A utility function to check if num has all 9s static bool isAll9( int [] num, int n) { for ( int i = 0; i < n; i++) if (num[i] != 9) return false ; return true ; } /* Utility that prints out an array on a line */ static void printarray( int []num) { for ( int i = 0; i < num.Length; i++) Console.Write(num[i]+ " " ); Console.Write( " " ); } // Driver code public static void Main() { int []num = { 9, 4, 1, 8, 7, 9, 7, 8, 3, 2, 2 }; generateNextPalindrome(num, num.Length); } } // This code is contributed by Smitha. |
PHP
<?php // PHP program to find next // smallest palindrome // Returns next palindrome // of a given number num[]. // This function is for // input type 2 and 3 function generateNextPalindromeUtil( $num , $n ) { $mid = (int)( $n / 2); // end of left side // is always 'mid -1' $i = $mid - 1; // Beginning of right // side depends if n // is odd or even $j = ( $n % 2 == 0) ? $mid : ( $mid + 1); // A bool variable to check // if copy of left side to // right is sufficient or not $leftsmaller = false; // Initially, ignore the // middle same digits while ( $i >= 0 && $num [ $i ] == $num [ $j ]) { $i --; $j ++; } // Find if the middle digit(s) // need to be incremented or // not (or copying left side // is not sufficient) if ( $i < 0 || $num [ $i ] < $num [ $j ]) { $leftsmaller = true; } // Copy the mirror // of left to tight while ( $i >= 0) { $num [ $j ++] = $num [ $i --]; } // Handle the case where // middle digit(s) must be // incremented. This part // of code is for CASE 1 // and CASE 2.2 if ( $leftsmaller ) { $carry = 1; // If there are odd digits, // then increment the middle // digit and store the carry if ( $n % 2 == 1) { $num [ $mid ] += 1; $carry = (int)( $num [ $mid ] / 10); $num [ $mid ] %= 10; } $i = $mid - 1; $j = ( $n % 2 == 0 ? $mid : $mid + 1); // Add 1 to the rightmost digit // of the left side, propagate // the carry towards MSB digit // and simultaneously copying // mirror of the left side to // the right side. while ( $i >= 0) { $num [ $i ] = $num [ $i ] + $carry ; $carry = (int)( $num [ $i ] / 10); $num [ $i ] %= 10; // copy mirror to right $num [ $j ] = $num [ $i ]; $i --; $j ++; } } return $num ; } // The function that prints // next palindrome of a given // number num[] with n digits. function generateNextPalindrome( $num , $n ) { echo "Next Palindrome is:" ; // Input type 1: All the // digits are 9, simply // o/p 1 followed by n-1 // 0's followed by 1. if (isAll9( $num , $n )) { echo "1" ; for ( $i = 0; $i < $n - 1; $i ++) echo "0" ; echo "1" ; } // Input type 2 and 3 else { $num = generateNextPalindromeUtil( $num , $n ); printarray( $num ); } } // A utility function to // check if num has all 9s function isAll9( $num , $n ) { for ( $i = 0; $i < $n ; $i ++) if ( $num [ $i ] != 9) return false; return true; } /* Utility that prints out an array on a line */ function printarray( $num ) { for ( $i = 0; $i < count ( $num ); $i ++) echo $num [ $i ]; echo "" ; } // Driver code $num = array (9, 4, 1, 8, 7, 9, 7, 8, 3, 2, 2); generateNextPalindrome( $num , count ( $num )); // This code is contributed by mits. ?> |
Javascript
<script> // JavaScript program to find next smallest // palindrome // Returns next palindrome of a given // number num. This function is for // input type 2 and 3 function generateNextPalindromeUtil(num , n) { var mid = parseInt(n / 2); // end of left side is always 'mid -1' var i = mid - 1; // Beginning of right side depends // if n is odd or even var j = (n % 2 == 0) ? mid : mid + 1; // A bool variable to check if copy of left // side to right // is sufficient or not leftsmaller = false ; // Initially, ignore the middle same digits while (i >= 0 && num[i] == num[j]) { i--; j++; } // Find if the middle digit(s) need to // be incremented or not (or copying left // side is not sufficient) if (i < 0 || num[i] < num[j]) { leftsmaller = true ; } // Copy the mirror of left to tight while (i >= 0) { num[j++] = num[i--]; } // Handle the case where middle digit(s) // must be incremented. This part of code // is for CASE 1 and CASE 2.2 if (leftsmaller) { var carry = 1; // If there are odd digits, then increment // the middle digit and store the carry if (n % 2 == 1) { num[mid] += 1; carry = parseInt(num[mid] / 10); num[mid] %= 10; } i = mid - 1; j = (n % 2 == 0 ? mid : mid + 1); // Add 1 to the rightmost digit of the left // side, propagate the carry towards MSB digit // and simultaneously copying mirror of the // left side to the right side. //when carry is zero no need to loop through till i>=0 while (i >= 0 && carry>0) { num[i] = num[i] + carry; carry = parseInt(num[i] / 10); num[i] %= 10; num[j] = num[i]; // copy mirror to right i--; j++; } } } // The function that prints next palindrome // of a given number num with n digits. function generateNextPalindrome(num , n) { document.write( "Next Palindrome is: <br>" ); // Input type 1: All the digits are 9, // simply o/p 1 followed by n-1 0's // followed by 1. if (isAll9(num, n)) { document.write( "1" ); for (i = 0; i < n - 1; i++) document.write( "0" ); document.write( "1" ); } // Input type 2 and 3 else { generateNextPalindromeUtil(num, n); printarray(num); } } // A utility function to check if num has all 9s function isAll9(num , n) { for (i = 0; i < n; i++) if (num[i] != 9) return false ; return true ; } /* Utility that prints out an array on a line */ function printarray(num) { for (i = 0; i < num.length; i++) document.write(num[i]+ "" ); } var num = [ 9, 4, 1, 8, 7, 9, 7, 8, 3, 2, 2 ]; generateNextPalindrome(num, num.length); // This code is contributed by 29AjayKumar </script> |
输出
Next palindrome is:9 4 1 8 8 0 8 8 1 4 9
如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。
© 版权声明
文章版权归作者所有，未经允许请勿转载。
THE END
![关于”PostgreSQL错误：关系[表]不存在“问题的原因和解决方案-yiteyi-C++库](https://www.yiteyi.com/wp-content/themes/zibll/img/thumbnail.svg)
