快乐数字

通过以下过程定义快乐数n。以n开头，将其替换为数字的平方和，然后重复该过程，直到n等于1，或者它在不包含1的循环中无休止地循环。 ，而那些不以1结尾的数字是不满意的数字。

前几个快乐数字是1，7，10，13，19，23，28，31，32，44，49，68，70，79，82，86，91，94，97，100

例子：

Input :  23
Output : Yes
Explanation : 
First Iteration:
22 + 32 = 4 + 9 = 13
Second Iteration:
12 + 32 = 1 + 9 = 10
Third Iteration:
12 + 02 = 1 + 0 = 1
Since we reach 1, 23 is Happy.

想法很简单，我们重复做数字平方和。在执行此操作时，我们使用哈希跟踪访问的号码。如果达到1，则返回true。否则，如果我们访问了一个号码，则返回false。

C++

// CPP program to check if a number
// is happy number
#include 
using namespace std;
  
// Returns sum of squares of digits
// of a number n. For example for n = 12
// it returns 1 + 4 = 5
int sumDigitSquare(int n)
{
    int sq = 0;
    while (n) {
        int digit = n % 10;
        sq += digit * digit;
        n = n / 10;
    }
    return sq;
}
  
// Returns true if n is Happy number
// else returns false.
bool isHappy(int n)
{
    // A set to store numbers during
    // repeated square sum process
    set s;
    s.insert(n);
  
    // Keep replacing n with sum of
    // squares of digits until we either
    // reach 1 or we endup in a cycle
    while (1) {
  
        // Number is Happy if we reach 1
        if (n == 1)
            return true;
  
        // Replace n with sum of squares
        // of digits
        n = sumDigitSquare(n);
  
        // If n is already visited, a cycle
        // is formed, means not Happy
        if (s.find(n) != s.end())
            return false;
  
        // Mark n as visited
        s.insert(n);
    }
  
    return false;
}
  
// Driver code
int main()
{
    int n = 23;
    if (isHappy(n))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

Java

// Java program to check if a number
// is happy number
import java.util.*;
  
class GFG 
{
  
// Returns sum of squares of digits
// of a number n. For example for n = 12
// it returns 1 + 4 = 5
static int sumDigitSquare(int n)
{
    int sq = 0;
    while (n > 0) 
    {
        int digit = n % 10;
        sq += digit * digit;
        n = n / 10;
    }
    return sq;
}
  
// Returns true if n is Happy number
// else returns false.
static boolean isHappy(int n)
{
    // A set to store numbers during
    // repeated square sum process
    HashSet s = new HashSet();
    s.add(n);
  
    // Keep replacing n with sum of
    // squares of digits until we either
    // reach 1 or we endup in a cycle
    while (true) 
    {
  
        // Number is Happy if we reach 1
        if (n == 1)
            return true;
  
        // Replace n with sum of squares
        // of digits
        n = sumDigitSquare(n);
  
        // If n is already visited, a cycle
        // is formed, means not Happy
          
        if ((s.contains(n) && n != (int)s.toArray()[ s.size()-1 ] ))
            return false;
  
        // Mark n as visited
        s.add(n);
    }
}
  
// Driver code
public static void main(String[] args) 
{
    int n = 23;
    if (isHappy(n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
/* This code contributed by PrinciRaj1992 */

Python 3

# python program to check if a number
# is happy number
  
  
# Returns sum of squares of digits
# of a number n. For example for n = 12
# it returns 1 + 4 = 5
def sumDigitSquare( n):
    sq = 0;
    while (n!=0):
        digit = n % 10
        sq += digit * digit
        n = n // 10
          
    return sq;
  
  
# Returns true if n is Happy number
# else returns false.
def isHappy(n):
    # A set to store numbers during
    # repeated square sum process
    s=set()
    s.add(n)
  
    # Keep replacing n with sum of
    # squares of digits until we either
    # reach 1 or we endup in a cycle
    while (True):
  
        # Number is Happy if we reach 1
        if (n == 1):
            return True;
  
        # Replace n with sum of squares
        # of digits
        n = sumDigitSquare(n)
  
        # If n is already visited, a cycle
        # is formed, means not Happy
        if n in s:
            return False
  
        # Mark n as visited
        s.add(n)
  
    return false;
  
  
# Driver code
  
n = 4
if (isHappy(n)):
    print("Yes") 
else:
    print("No")

C#

// C# program to check if a number
// is happy number
using System; 
using System.Collections.Generic; 
  
class GFG 
{
  
// Returns sum of squares of digits
// of a number n. For example for n = 12
// it returns 1 + 4 = 5
static int sumDigitSquare(int n)
{
    int sq = 0;
    while (n > 0) 
    {
        int digit = n % 10;
        sq += digit * digit;
        n = n / 10;
    }
    return sq;
}
  
// Returns true if n is Happy number
// else returns false.
static bool isHappy(int n)
{
    // A set to store numbers during
    // repeated square sum process
    HashSet s = new HashSet();
    s.Add(n);
  
    // Keep replacing n with sum of
    // squares of digits until we either
    // reach 1 or we endup in a cycle
    while (true) 
    {
  
        // Number is Happy if we reach 1
        if (n == 1)
            return true;
  
        // Replace n with sum of squares
        // of digits
        n = sumDigitSquare(n);
  
        // If n is already visited, a cycle
        // is formed, means not Happy
        if (s.Contains(n))
            return false;
  
        // Mark n as visited
        s.Add(n);
    }
}
  
// Driver code
public static void Main() 
{
    int n = 23;
    if (isHappy(n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
  
// This code contributed by Rajput-Ji

输出：

Yes

一个重要的发现是，循环始终包含4 。因此，我们不需要跟踪所有数字。我们可以简单地检查4。

C++

// A space optimized CPP program to check 
// if a number is happy number
#include 
using namespace std;
  
// Returns sum of squares of digits
// of a number n. For example for n = 12
// it returns 1 + 4 = 5
int sumDigitSquare(int n)
{
    int sq = 0;
    while (n) {
        int digit = n % 10;
        sq += digit * digit;
        n = n / 10;
    }
    return sq;
}
  
// Returns true if n is Happy number
// else returns false.
bool isHappy(int n)
{
    // Keep replacing n with sum of
    // squares of digits until we either
    // reach 1 or we end up in a cycle
    while (1) {
  
        // Number is Happy if we reach 1
        if (n == 1)
            return true;
  
        // Replace n with sum of squares
        // of digits
        n = sumDigitSquare(n);
  
        // Number is not Happy if we reach 4
        if (n == 4)
            return false;
    }
  
    return false;
}
  
// Driver code
int main()
{
    int n = 23;
    if (isHappy(n))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

Java

// A space optimized Java 
// program to check if a
// number is happy number
import java.io.*;
  
class GFG {
      
    // Returns sum of squares of 
    // digits of a number n. For 
    // example for n = 12 it 
    // returns 1 + 4 = 5
    static int sumDigitSquare(int n)
    {
        int sq = 0;
        while (n != 0) 
        {
            int digit = n % 10;
            sq += digit * digit;
            n = n / 10;
        }
        return sq;
    }
      
    // Returns true if n is Happy 
    // number else returns false.
    static boolean isHappy(int n)
    {
        // Keep replacing n with 
        // sum of squares of digits 
        // until we either reach 1 
        // or we end up in a cycle
        while (true) 
        {
      
            // Number is Happy if 
            // we reach 1
            if (n == 1)
                return true;
      
            // Replace n with sum 
            // of squares of digits
            n = sumDigitSquare(n);
      
            // Number is not Happy 
            // if we reach 4
            if (n == 4)
                return false;
        }
      
    }
      
    // Driver code
    public static void main(String args[])
    {
        int n = 23;
        if (isHappy(n))
            System.out.println("Yes");
        else
            System.out.println("No" );
    }
}
  
/*This code is contributed by Nikita tiwari.*/

Python3

# A space optimized Python3 program to 
# check  if a number is happy number
  
  
# Returns sum of squares of 
# digits of a number n. For 
# example for n = 12 it 
# returns 1 + 4 = 5
def sumDigitSquare(n) :
    sq = 0
    while (n) :
        digit = n % 10
        sq = sq + digit * digit
        n = n // 10
      
    return sq
  
  
# Returns true if n 
# is Happy number else 
# returns false.
def isHappy(n) :
  
    # Keep replacing n with 
    # sum of squares of digits
    # until we either reach 1 
    # or we end up in a cycle
    while (1) :
  
        # Number is Happy if 
        # we reach 1
        if (n == 1) :
            return True
  
        # Replace n with sum of 
        # squares of digits
        n = sumDigitSquare(n)
  
        # Number is not Happy 
        # if we reach 4
        if (n == 4) :
            return False
      
    return False
  
# Driver code
n = 23
if (isHappy(n)) :
    print("Yes")
else :
    print("No")
      
  
# This code is contributed 
# by Nikita tiwari.

C＃

// A space optimized C# 
// program to check if a
// number is happy number
using System;
  
class GFG
{
      
// Returns sum of squares of 
// digits of a number n. For 
// example for n = 12 it 
// returns 1 + 4 = 5
static int sumDigitSquare(int n)
{
    int sq = 0;
    while (n != 0) 
    {
        int digit = n % 10;
        sq += digit * digit;
        n = n / 10;
    }
    return sq;
}
  
// Returns true if n is Happy 
// number else returns false.
static bool isHappy(int n)
{
    // Keep replacing n with 
    // sum of squares of digits 
    // until we either reach 1 
    // or we end up in a cycle
    while (true) 
    {
  
        // Number is Happy if 
        // we reach 1
        if (n == 1)
            return true;
  
        // Replace n with sum 
        // of squares of digits
        n = sumDigitSquare(n);
  
        // Number is not Happy 
        // if we reach 4
        if (n == 4)
            return false;
    }
  
}
  
// Driver code
static public void Main ()
{
    int n = 23;
    if (isHappy(n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No" );
}
}
  
// This code is contributed by ajit 

输出：

Yes