LeetCode#202_Cover Photo

LeetCode #202 Happy Number Solution & Explanation

Klook.com
Post Views: 307

LeetCode Problem

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.

Solution

In the first solution, we use an array to store the number that have used ( numbers that aren’t Happy)

And in the next iteration, if any numbers repeat again, we can return False.

C# Solution

Solution1

public class Solution {
    public bool IsHappy(int n) {
        
        List<int>used = new List<int>();

        while(n>0)
        {
            int tmp = 0;
            while(n>0)
            {
                int i = n%10;

                tmp += i*i;
                n = n/10;
            }

            if(used.Contains(tmp))
            {
                return false;
            }
            else
            {
                used.Add(tmp);    
            }
            

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

            n = tmp;

        }
        
        return false;
    }
}

Java Solution

Solution1

class Solution {
    public boolean isHappy(int n) {
        List<Integer>used = new ArrayList<Integer>();

        while(n>0)
        {
            int tmp = 0;
            while(n>0)
            {
                int i = n%10;

                tmp += i*i;
                n = n/10;
            }

            if(used.contains(tmp))
            {
                return false;
            }
            else
            {
                used.add(tmp);    
            }
            

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

            n = tmp;

        }
        
        return false; 
    }
}

Runtime : 1ms、1ms、3ms

Python3 Solution

Solution1

class Solution:
    def isHappy(self, n: int) -> bool:
        used = []

        while(n>0):
            tmp = 0
            while(n>0):
                i = n%10

                tmp += i*i
                n = n//10

            if(tmp in used):
                return False
            else:
                used.append(tmp)

            if(tmp==1):
                return True
            
            n = tmp

        return False

Python’s // give result to integer

JavaScript Solution

Solution1

/**
 * @param {number} n
 * @return {boolean}
 */
var isHappy = function(n) {
    let used = new Set();

    while(n>0)
    {
        let tmp = 0;
        while(n>0)
        {
            let i = n%10;

            tmp += i*i;
            n = Math.floor(n/10);
        }

        if(used.has(tmp))
        {
            return false;
        }
        else
        {
            used.add(tmp);    
        }
        

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

        n = tmp;

    }
    
    return false; 
};

Runtime : 69ms、69ms、66ms ≈ Faster than 80~90%

You can also use array if you want, but remember use array’s includes() to replace set’s has()

Conclusion

🧡If my solution helps, that is my honor!

🧡You can support me by clicking some ad, and that will make me Happy

✅If you got any problem about the explanation, please feel free to let me know

Cover Photo – Photo by Count Chris on Unsplash

The problem link : Happy Number – LeetCode

Random post

Related Posts

Leave a Reply

Your email address will not be published. Required fields are marked *