Question

Formatted question description: https://leetcode.ca/all/202.html

 202	Happy Number

 Write an algorithm to determine if a number 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,
    and 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 numbers.

 Example: 19 is a happy number

 1^2 + 9^2 = 82
 8^2 + 2^2 = 68
 6^2 + 8^2 = 100
 1^2 + 0^2 + 0^2 = 1

Algorithm

The example 19 given in the title is a happy number, so let’s look at a situation that is not a happy number. For example, the number 11 has the following calculation process:

1^2 + 1^2 = 2 2^2 = 4 4^2 = 16 1^2 + 6^2 = 37 3^2 + 7^2 = 58 5^2 + 8^2 = 89 8^2 + 9^2 = 145 1^2 + 4^2 + 5^2 = 42 4^2 + 2^2 = 20 2^2 + 0^2 = 4

At the end of the calculation, the number 4 appears again, then the following numbers will repeat the previous order. This cycle does not contain 1, then the number 11 is not a happy number. After discovering the pattern, we can consider how to use code it.

Use HashSet to record all the numbers that have appeared, and then every time a new number appears, look for it in HashSet to see if it exists,

  • if it does not exist, add it to the table,
  • if it exists, jump out of the loop, and determine whether the number is 1,
    • if it is 1. Return true,
    • return false if not 1

Code

Java

  • 
        class Solution {
            public boolean isHappy(int n) {
    
                HashSet<Integer> set = new HashSet<Integer>();
    
                while (!set.contains(n)) {
                    set.add(n);
    
                    n = getSum(n);
    
                    if (n == 1) {
                        return true;
                    }
                }
    
                return false;
            }
    
            public int getSum(int current) {
                int sum = 0;
    
                while (current > 0) {
                    sum += Math.pow(current % 10, 2);
                    current /= 10;
                }
    
                return sum;
            }
        }
    
        class Solution_iteration {
            public boolean isHappy(int n) {
                HashSet<Integer> st = new HashSet<Integer>();
                while (n != 1) {
                    int sum = 0;
                    while (n > 0) {
                        sum += (n % 10) * (n % 10);
                        n /= 10;
                    }
                    n = sum;
                    if (st.contains(n)) {
                        break;
                    }
                    st.add(n);
                }
    
                return n == 1;
            }
        }
    
        // if find 4 then unhappy
        class Solution_math {
            public boolean isHappy(int n) {
                while (n != 1 && n != 4) {
                    int sum = 0;
                    while (n > 0) {
                        sum += (n % 10) * (n % 10);
                        n /= 10;
                    }
                    n = sum;
                }
                return n == 1;
    
            }
        }
    }
    
    
  • // OJ: https://leetcode.com/problems/happy-number/
    // Time: O(1)
    // Space: O(1)
    class Solution {
    public:
      bool isHappy(int n) {
        unordered_set<int> s;
        while (n != 1 && !s.count(n)) {
          s.insert(n);
          int next = 0;
          while (n) {
            next += pow(n % 10, 2);
            n /= 10;
          }
          n = next;
        }
        return n == 1;
      }
    };
    
  • class Solution(object):
      def isHappy(self, n):
        """
        :type n: int
        :rtype: bool
        """
        record = {}
        sq_sum = 0
        while n != 1:
          sq_sum = 0
          sub_num = n
          while sub_num > 0:
            sq_sum += (sub_num % 10) * (sub_num % 10)
            sub_num /= 10
          if sq_sum in record:
            return False
          else:
            record[sq_sum] = 1
          n = sq_sum
        return True
    
    

All Problems

All Solutions