Glad nummer
Prøv det på GfG Practice 
Et tal kaldes lykkeligt, hvis det fører til 1 efter en sekvens af trin, hvor hvert trinnummer erstattes af summen af kvadrater af dets ciffer, det vil sige, hvis vi starter med lykkeligt tal og bliver ved med at erstatte det med cifre i kvadratsum, når vi 1.
Eksempler:
Input: n = 19
Output: True
19 is Happy Number
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1
As we reached to 1 19 is a Happy Number.
Input: n = 20
Output: FalseEt tal vil ikke være et lykkeligt tal, når det laver en løkke i sin rækkefølge, dvs. det rører ved et tal i rækkefølge, som allerede er blevet rørt. Så for at kontrollere, om et tal er lykkeligt eller ej, kan vi beholde et sæt, hvis det samme tal opstår igen, markerer vi resultatet som ikke lykkeligt. En simpel funktion på ovenstående tilgang kan skrives som nedenfor -
C++Java// method return true if n is Happy Number int numSquareSum ( int n ) { int num = 0 ; while ( n != 0 ) { int digit = n % 10 ; num += digit * digit ; n /= 10 ; } return num ; } int isHappyNumber ( int n ) { set < int > st ; while ( 1 ) { n = numSquareSum ( n ); if ( n == 1 ) return true ; if ( st . find ( n ) != st . end ()) return false ; st . insert ( n ); } }Python// method return true if n is Happy Number public static int numSquareSum ( int n ) { int num = 0 ; while ( n != 0 ) { int digit = n % 10 ; num += digit * digit ; n /= 10 ; } return num ; } static boolean isHappyNumber ( int n ) { HashSet < Integer > st = new HashSet <> (); while ( true ) { n = numSquareSum ( n ); if ( n == 1 ) return true ; if ( st . contains ( n )) return false ; st . add ( n ); } } // This code is contributed by Princi SinghC## method return true if n is Happy Number def numSquareSum ( n ): num = 0 while ( n ): digit = n % 10 num = num + digit * digit n = n // 10 return num def isHappyNumber ( n ): st = set () while ( 1 ): n = numSquareSum ( n ) if ( n == 1 ): return True if n not in st : return False st . insert ( n )JavaScript// Method return true if n is Happy Number static int numSquareSum ( int n ) { int num = 0 ; while ( n != 0 ) { int digit = n % 10 ; num += digit * digit ; n /= 10 ; } return num ; } static int isHappyNumber ( int n ) { HashSet < int > st = new HashSet <> (); while ( 1 ) { n = numSquareSum ( n ); if ( n == 1 ) return true ; if ( st . Contains ( n )) return false ; st . Add ( n ); } } // This code is contributed by 29AjayKumarKompleksitetsanalyse:
Tidskompleksitet: O(n*log(n)).
Hjælpeplads: O(n) siden brug af ekstra sæt til opbevaringVi kan løse dette problem uden at bruge ekstra plads, og den teknik kan også bruges i nogle andre lignende problemer. Hvis vi behandler hvert tal som en knude og erstatning med kvadratsumciffer som et link, er dette problem det samme som finde en løkke i en linkliste :
Så som en foreslået løsning fra ovenstående link vil vi holde to tal langsomme og hurtige, begge initialisere fra et givet tal langsomt erstattes et trin ad gangen og hurtigt erstattes to trin ad gangen. Hvis de mødes ved 1, så er det angivne antal Happy Number ellers ikke.
C++C// C++ program to check a number is a Happy number or not #includeusing namespace std ; // Utility method to return sum of square of digit of n int numSquareSum ( int n ) { int squareSum = 0 ; while ( n ) { squareSum += ( n % 10 ) * ( n % 10 ); n /= 10 ; } return squareSum ; } // method return true if n is Happy number bool isHappynumber ( int n ) { int slow fast ; // initialize slow and fast by n slow = fast = n ; do { // move slow number by one iteration slow = numSquareSum ( slow ); // move fast number by two iteration fast = numSquareSum ( numSquareSum ( fast )); } while ( slow != fast ); // if both number meet at 1 then return true return ( slow == 1 ); } // Driver code to test above methods int main () { int n = 13 ; if ( isHappynumber ( n )) cout < < n < < ' is a Happy number n ' ; else cout < < n < < ' is not a Happy number n ' ; } // This code is contributed by divyeshrabadiya07 Java// C program to check a number is a Happy number or not #include#include // Utility method to return sum of square of digit of n int numSquareSum ( int n ) { int squareSum = 0 ; while ( n ) { squareSum += ( n % 10 ) * ( n % 10 ); n /= 10 ; } return squareSum ; } // method return true if n is Happy number bool isHappynumber ( int n ) { int slow fast ; // initialize slow and fast by n slow = fast = n ; do { // move slow number by one iteration slow = numSquareSum ( slow ); // move fast number by two iteration fast = numSquareSum ( numSquareSum ( fast )); } while ( slow != fast ); // if both number meet at 1 then return true return ( slow == 1 ); } // Driver code to test above methods int main () { int n = 13 ; if ( isHappynumber ( n )) printf ( '%d is a Happy number n ' n ); else printf ( '%d is not a Happy number n ' n ); } // This code is contributed by Sania Kumari Gupta // (kriSania804) Python// Java program to check a number is a Happy // number or not class GFG { // Utility method to return sum of square of // digit of n static int numSquareSum ( int n ) { int squareSum = 0 ; while ( n != 0 ) { squareSum += ( n % 10 ) * ( n % 10 ); n /= 10 ; } return squareSum ; } // method return true if n is Happy number static boolean isHappynumber ( int n ) { int slow fast ; // initialize slow and fast by n slow = fast = n ; do { // move slow number // by one iteration slow = numSquareSum ( slow ); // move fast number // by two iteration fast = numSquareSum ( numSquareSum ( fast )); } while ( slow != fast ); // if both number meet at 1 // then return true return ( slow == 1 ); } // Driver code to test above methods public static void main ( String [] args ) { int n = 13 ; if ( isHappynumber ( n )) System . out . println ( n + ' is a Happy number' ); else System . out . println ( n + ' is not a Happy number' ); } }C## Python3 program to check if a number is a Happy number or not # Utility method to return the sum of squares of digits of n def num_square_sum ( n ): square_sum = 0 while n : square_sum += ( n % 10 ) ** 2 n //= 10 return square_sum # Method returns True if n is a Happy number def is_happy_number ( n ): # Initialize slow and fast pointers slow = n fast = n while True : # Move slow pointer by one iteration slow = num_square_sum ( slow ) # Move fast pointer by two iterations fast = num_square_sum ( num_square_sum ( fast )) if slow != fast : continue else : break # If both pointers meet at 1 then return True return slow == 1 # Driver Code n = 13 if is_happy_number ( n ): print ( n 'is a Happy number' ) else : print ( n 'is not a Happy number' )JavaScript// C# program to check a number // is a Happy number or not using System ; class GFG { // Utility method to return // sum of square of digit of n static int numSquareSum ( int n ) { int squareSum = 0 ; while ( n != 0 ) { squareSum += ( n % 10 ) * ( n % 10 ); n /= 10 ; } return squareSum ; } // method return true if // n is Happy number static bool isHappynumber ( int n ) { int slow fast ; // initialize slow and // fast by n slow = fast = n ; do { // move slow number // by one iteration slow = numSquareSum ( slow ); // move fast number // by two iteration fast = numSquareSum ( numSquareSum ( fast )); } while ( slow != fast ); // if both number meet at 1 // then return true return ( slow == 1 ); } // Driver code public static void Main () { int n = 13 ; if ( isHappynumber ( n )) Console . WriteLine ( n + ' is a Happy number' ); else Console . WriteLine ( n + ' is not a Happy number' ); } } // This code is contributed by anuj_67.PHP< script > // Javascript program to check a number is a Happy // number or not // Utility method to return sum of square of // digit of n function numSquareSum ( n ) { var squareSum = 0 ; while ( n != 0 ) { squareSum += ( n % 10 ) * ( n % 10 ); n = parseInt ( n / 10 ); } return squareSum ; } // method return true if n is Happy number function isHappynumber ( n ) { var slow fast ; // initialize slow and fast by n slow = fast = n ; do { // move slow number // by one iteration slow = numSquareSum ( slow ); // move fast number // by two iteration fast = numSquareSum ( numSquareSum ( fast )); } while ( slow != fast ); // if both number meet at 1 // then return true return ( slow == 1 ); } // Driver code to test above methods var n = 13 ; if ( isHappynumber ( n )) document . write ( n + ' is a Happy number' ); else document . write ( n + ' is not a Happy number' ); // This code contributed by Princi Singh < /script>Output:
13 is a Happy NumberKompleksitetsanalyse:
Tidskompleksitet: O(n*log(n)).
Hjælpeplads: O(1).C++
En anden tilgang til at løse dette problem uden ekstra plads.
Et tal kan ikke være et lykkeligt tal hvis summen af kvadratet af cifre på noget trin er et enkeltcifret tal undtagen 1 eller 7 . Dette skyldes, at 1 og 7 er de eneste encifrede lykkelige tal. Ved at bruge disse oplysninger kan vi udvikle en tilgang som vist i koden nedenfor -C// C++ program to check if a number is a Happy number or // not. #includeusing namespace std ; // Method - returns true if the input is a happy number else // returns false bool isHappynumber ( int n ) { int sum = n x = n ; // This loop executes till the sum of square of digits // obtained is not a single digit number while ( sum > 9 ) { sum = 0 ; // This loop finds the sum of square of digits while ( x > 0 ) { int d = x % 10 ; sum += d * d ; x /= 10 ; } x = sum ; } if ( sum == 7 || sum == 1 ) return true ; return false ; } int main () { int n = 13 ; if ( isHappynumber ( n )) cout < < n < < ' is a Happy number' ; else cout < < n < < ' is not a Happy number' ; return 0 ; } // This code is contributed by Sania Kumari Gupta Java// C program to check if a number is a Happy number or // not. #include#include // Method - returns true if the input is a happy number else // returns false bool isHappynumber ( int n ) { int sum = n x = n ; // This loop executes till the sum of square of digits // obtained is not a single digit number while ( sum > 9 ) { sum = 0 ; // This loop finds the sum of square of digits while ( x > 0 ) { int d = x % 10 ; sum += d * d ; x /= 10 ; } x = sum ; } if ( sum == 7 || sum == 1 ) return true ; return false ; } int main () { int n = 13 ; if ( isHappynumber ( n )) printf ( '%d is a Happy number' n ); else printf ( '%d is not a Happy number' n ); return 0 ; } // This code is contributed by Sania Kumari Gupta Python// This code is contributed by Vansh Sodhi. // Java program to check if a number is a Happy number or // not. class GFG { // method - returns true if the input is a happy // number else returns false static boolean isHappynumber ( int n ) { int sum = n x = n ; // this loop executes till the sum of square of // digits obtained is not a single digit number while ( sum > 9 ) { sum = 0 ; // this loop finds the sum of square of digits while ( x > 0 ) { int d = x % 10 ; sum += d * d ; x /= 10 ; } x = sum ; } if ( sum == 1 || sum == 7 ) return true ; return false ; } // Driver code public static void main ( String [] args ) { int n = 13 ; if ( isHappynumber ( n )) System . out . println ( n + ' is a Happy number' ); else System . out . println ( n + ' is not a Happy number' ); } }C## Python3 program to check if a number is a Happy number or not. # Method - returns true if the input is # a happy number else returns false def isHappynumber ( n ): Sum x = n n # This loop executes till the sum # of square of digits obtained is # not a single digit number while Sum > 9 : Sum = 0 # This loop finds the sum of # square of digits while x > 0 : d = x % 10 Sum += d * d x = int ( x / 10 ) x = Sum if Sum == 1 or Sum == 7 : return True return False n = 13 if isHappynumber ( n ): print ( n 'is a Happy number' ) else : print ( n 'is not a Happy number' ) # This code is contributed by mukesh07.JavaScript// C# program to check if a number // is a Happy number or not. using System ; class GFG { // Method - returns true if the input is // a happy number else returns false static bool isHappynumber ( int n ) { int sum = n x = n ; // This loop executes till the sum // of square of digits obtained is // not a single digit number while ( sum > 9 ) { sum = 0 ; // This loop finds the sum of // square of digits while ( x > 0 ) { int d = x % 10 ; sum += d * d ; x /= 10 ; } x = sum ; } if ( sum == 1 || sum == 7 ) return true ; return false ; } // Driver code public static void Main ( String [] args ) { int n = 13 ; if ( isHappynumber ( n )) Console . WriteLine ( n + ' is a Happy number' ); else Console . WriteLine ( n + ' is not a Happy number' ); } } // This code is contributed by 29AjayKumar
Produktion13 is a Happy numberKompleksitetsanalyse:
Tidskompleksitet: O(n*log(n)).
Hjælpeplads: O(1).Se din artikel, der vises på GeeksforGeeks' hovedside, og hjælp andre nørder.
Opret quiz
