Fuck LeetCode  >  Hash Table  >  202. Happy Number  >  Solved in Java, Python, JavaScript, C#, Go, C++, Ruby  >  Repost or Contribute

    LeetCode link: 202. Happy Number, difficulty: Easy.

    Write an algorithm to determine if a number n is happy.

    A happy number is a number defined by the following process:

    • Starting with any positive integer, replace the number by the sum of the squares of its digits.
    • Repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1.
    • Those numbers for which this process ends in 1 are happy.

    Return true if n is a happy number, and false if not.

    Example 1:

    Input: n = 19

    Output: true

    Explanation:

    12 + 92 = 82
    82 + 22 = 68
    62 + 82 = 100
    12 + 02 + 02 = 1

    Example 2:

    Input: n = 2

    Output: false

    Constraints:

    • 1 <= n <= 2^31 - 1

    Intuition
    PreviousNext

    1. It is more convenient to call isHappy(n) recursively. You only need to generate a new n as a parameter each time.
    2. If n has already appeared, it means that the loop has been entered, and return false. You can use Set to save the n that has appeared.
    3. Go is the iterative solution, other languages are the recursive solution.

    Steps

    1. Generate a new n as the isHappy(n) parameter.

      let sum = 0

for (const digit of n.toString()) {
  sum += Math.pow(Number(digit), 2)
}

// omitted code

return isHappy(sum)

    2. If n has already appeared, it means that the loop has been entered, and return false. You can use Set to save the n that has appeared.

      var isHappy = function (n, appearedNums) { // 0
  appearedNums ||= new Set() // 1
  let sum = 0

  for (const digit of n.toString()) {
    sum += Math.pow(Number(digit), 2)
  }

  if (sum == 1) {
    return true
  }

  if (appearedNums.has(sum)) { // 2
    return false
  }

  appearedNums.add(sum) // 3

  return isHappy(sum, appearedNums)
};

    Complexity

    Time complexity

    O(log N)

    Space complexity

    O(log N)

    Solution in languages: Java Python JavaScript C# Go C++ Ruby

    Java # 

    class Solution {
    private HashSet<Integer> appearedNums = new HashSet<>();

    public boolean isHappy(int n) {
        appearedNums.add(n);

        var sum = 0;

        for (var digit : String.valueOf(n).toCharArray()) {
            sum += Math.pow(digit - '0', 2);
        }

        if (sum == 1) {
            return true;
        }

        if (appearedNums.contains(sum)) {
            return false;
        }

        return isHappy(sum);
    }
}

    Python # 

    class Solution:
    def __init__(self):
        self.appeared_nums = set()

    def isHappy(self, n: int) -> bool:
        self.appeared_nums.add(n)

        number = sum([int(digit) ** 2 for digit in str(n)])

        if number == 1:
            return True

        if number in self.appeared_nums:
            return False

        return self.isHappy(number)

    JavaScript # 

    var isHappy = function (n, appearedNums) {
  appearedNums ||= new Set()
  let sum = 0

  for (const digit of n.toString()) {
    sum += Math.pow(Number(digit), 2)
  }

  if (sum == 1) {
    return true
  }

  if (appearedNums.has(sum)) {
    return false
  }

  appearedNums.add(sum)

  return isHappy(sum, appearedNums)
};

    C# # 

    public class Solution
{
    HashSet<int> appearedNums = new HashSet<int>();

    public bool IsHappy(int n)
    {
        appearedNums.Add(n);

        int sum = 0;

        foreach (char digit in n.ToString().ToCharArray())
            sum += (int)Math.Pow(digit - '0', 2);

        if (sum == 1)
            return true;

        if (appearedNums.Contains(sum))
            return false;

        return IsHappy(sum);
    }
}

    Go # 

    func isHappy(n int) bool {
    // Use map to track seen numbers
    seen := make(map[int]bool)

    for n != 1 && !seen[n] {
        seen[n] = true
        n = sumOfSquaredDigits(n)
    }

    return n == 1
}

func sumOfSquaredDigits(n int) int {
    sum := 0
    nStr := strconv.Itoa(n)
    for i := 0; i < len(nStr); i++ {
        digit := int(nStr[i] - '0')
        sum += digit * digit
    }
    return sum
}

    C++ # 

    class Solution {
public:
    bool isHappy(int n) {
        if (n == 1) {
            return true;
        }

        if (appeared_nums_.count(n)) {
            return false;
        }

        appeared_nums_.insert(n);

        return isHappy(getSum(n));
    }

private:
    unordered_set<int> appeared_nums_;

    int getSum(int n) {
        string n_str = to_string(n);
        int sum = 0;
        for (char digit : n_str) {
            sum += (digit - '0') * (digit - '0');
        }
        return sum;
    }
};

    Ruby # 

    def is_happy(n, appeared_nums = Set.new)
  return true if n == 1
  return false if appeared_nums.include?(n)

  appeared_nums.add(n)
  sum = n.to_s.chars.map { |digit| digit.to_i ** 2 }.sum

  is_happy(sum, appeared_nums)
end
    Solution in languages: Java Python JavaScript C# Go C++ Ruby

    Other languages

    Welcome to contribute code to our GitHub repository, thanks! The location of this solution is 202. Happy Number.
    PreviousNext