202. Happy Number
1. Description
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.
2. Example
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
3. Constraints
- 1 <= n <= $2^{31} - 1$
4. Solutions
Hash Table
Time complexity: O(logn)
Space complexity: O(logn)
class Solution {
public:
bool isHappy(const int n) {
unordered_map<int, bool> num_occur;
for (int sum = n; sum != 1; ) {
int next_sum = 0;
while (sum > 0) {
int digit = sum % 10;
next_sum += digit * digit;
sum /= 10;
}
sum = next_sum;
if (num_occur[sum]) {
return false;
}
num_occur[sum] = true;
}
return true;
}
};
Two Pointers
Time complexity: O(logn)
Space complexity: O(logn)
class Solution {
public:
bool isHappy(int n) {
int slow_number = n;
int fast_number = get_next_number_(n);
while (fast_number != 1 && slow_number != fast_number) {
slow_number = get_next_number_(slow_number);
fast_number = get_next_number_(get_next_number_(fast_number));
}
return fast_number == 1;
}
private:
int get_next_number_(int n) {
int digit_sum = 0;
for (; n > 0; n /= 10) {
int digit = n % 10;
digit_sum += digit * digit;
}
return digit_sum;
}
};