Happy Number 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 <= 231 - 1
Happy Number Solutions
✅Time: O(log n)
✅Space: O(n)
C++
class Solution {
public:
bool isHappy(int n) {
auto squaredSum = [&](int n) {
int sum = 0;
while (n) {
sum += pow(n % 10, 2);
n /= 10;
}
return sum;
};
int slow = squaredSum(n);
int fast = squaredSum(squaredSum(n));
while (slow != fast) {
slow = squaredSum(slow);
fast = squaredSum(squaredSum(fast));
}
return slow == 1;
}
};
Java
class Solution {
public boolean isHappy(int n) {
int slow = squaredSum(n);
int fast = squaredSum(squaredSum(n));
while (slow != fast) {
slow = squaredSum(slow);
fast = squaredSum(squaredSum(fast));
}
return slow == 1;
}
private int squaredSum(int n) {
int sum = 0;
while (n > 0) {
sum += Math.pow(n % 10, 2);
n /= 10;
}
return sum;
}
}
Python
class Solution:
def isHappy(self, n: int) -> bool:
def squaredSum(n: int) -> bool:
sum = 0
while n:
sum += pow(n % 10, 2)
n //= 10
return sum
slow = squaredSum(n)
fast = squaredSum(squaredSum(n))
while slow != fast:
slow = squaredSum(slow)
fast = squaredSum(squaredSum(fast))
return slow == 1
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