Skaičius vadinamas laimingu, jei po žingsnių sekos jis veda į 1, kai kiekvienas žingsnio numeris pakeičiamas jo skaitmens kvadratų suma, ty jei pradedame nuo laimingo skaičiaus ir nuolat keičiame jį skaitmenimis, kurių kvadratinė suma pasiekiame 1.
Pavyzdžiai:
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: False
Skaičius nebus laimingas skaičius, kai jis savo sekoje padaro kilpą, ty paliečia skaičių iš eilės, kuri jau buvo paliesta. Taigi norėdami patikrinti, ar skaičius yra laimingas, ar ne, galime išlaikyti rinkinį, jei tas pats skaičius pasikartoja, pažymime rezultatą kaip nelaimingą. Paprasta aukščiau pateikto metodo funkcija gali būti parašyta taip, kaip nurodyta toliau –
C++// 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); } }
// 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 Singh
# 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)
// 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 29AjayKumar
<script> // method return true if n is Happy Number function numSquareSum(n) { let num = 0; while (n !== 0) { let digit = n % 10; num += digit * digit; n = Math.floor(n / 10); } return num; } let st = new Set(); while (1) { n = numSquareSum(n); if (n == 1) return true; if (st.has(n)) return false; st.add(n); } } //This code is contributed by Mayank Tyagi </script>
Sudėtingumo analizė:
Laiko sudėtingumas: O(n*log(n)).
Pagalbinė erdvė: O(n), nes saugojimui naudojamas papildomas rinkinys
Šią problemą galime išspręsti nenaudodami papildomos vietos, o ši technika gali būti naudojama ir kai kuriose kitose panašiose problemose. Jei kiekvieną skaičių traktuosime kaip mazgą, o pakeitimą kvadratinės sumos skaitmeniu kaip nuorodą, tada ši problema yra tokia pati kaip kilpos radimas nuorodų sąraše :
Taigi, kaip siūlomas sprendimas iš pirmiau pateiktos nuorodos, mes išlaikysime du skaičius lėtus ir greitus, tiek inicijavimas iš nurodyto skaičiaus lėtas keičiamas po vieną žingsnį, o greitas keičiamas dviem žingsniais vienu metu. Jei jie susitinka 1, tada nurodytas skaičius yra laimingas skaičius, kitaip ne.
C++// C++ program to check a number is a Happy number or not #include
// C program to check a number is a Happy number or not #include
// 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'); } }
# 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')
// 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.
<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>
// PHP 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) { $squareSum = 0; while ($n) { $squareSum += ($n % 10) * ($n % 10); $n /= 10; } return $squareSum; } // method return true if // n is Happy number function isHappynumber( $n) { $slow; $fast; // initialize slow // and fast by n $slow = $n; $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 $n = 13; if (isHappynumber($n)) echo $n ' is a Happy numbern'; else echo n ' is not a Happy numbern'; // This code is contributed by anuj_67. ?> Išvestis:
13 is a Happy NumberSudėtingumo analizė:
Laiko sudėtingumas: O(n*log(n)).
Pagalbinė erdvė: O(1).
Kitas būdas išspręsti šią problemą nenaudojant papildomos vietos.
Skaičius negali būti laimingas skaičius jei bet kuriame žingsnyje gautų skaitmenų kvadratų suma yra vienaženklis skaičius, išskyrus 1 arba 7 . Taip yra todėl, kad 1 ir 7 yra vieninteliai vienženkliai laimingi skaičiai. Naudodami šią informaciją galime sukurti metodą, kaip parodyta toliau pateiktame kode -
// C++ program to check if a number is a Happy number or // not. #include
// C program to check if a number is a Happy number or // not. #include
// 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'); } }
# 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.
// 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
<script> // This code is contributed by Vansh Sodhi. // javascript 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 function isHappynumber(n) { var 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) { var d = x % 10; sum += d * d; x /= 10; } x = sum; } if(sum == 1 || sum == 7) return true; return false; } // Driver code var n = 13; if (isHappynumber(n)) document.write(n + ' is a Happy number'); else document.write(n + ' is not a Happy number'); // This code is contributed by 29AjayKumar </script>
Išvestis
13 is a Happy number
Sudėtingumo analizė:
Laiko sudėtingumas: O(n*log(n)).
Pagalbinė erdvė: O(1).
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