202. Happy Number

https://leetcode.com/problems/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:

Input: 19
Output: true
Explanation: 
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1

Thoughts

给定一个数，对每个数字做平方后求和形成新数，继续对数字求平方和，依次类推问能否得到1。按照要求生成新的number, 不收敛意味着之前遇到过，用hash set。时间复杂度O(lgN)，证明起来比较复杂。

Code

class Solution:
    def isHappy(self, n: int) -> bool:
        v = set()
        while n not in v:
            if n == 1:
                return True
            v.add(n)
            n = sum(int(c) ** 2 for c in str(n))
        return False

Analysis

空间复杂度O(n).

Ver.2

检测环想到Floyd算法。

class Solution {
    private int newNum(int n) {
        int res = 0;
        while (n / 10 != 0) {
            res += Math.pow(n % 10, 2); 
            n = n / 10;
        }
        if (n != 0) {
            res += Math.pow(n, 2);
        }
        return res;
    }

    public boolean isHappy(int n) {
        int slow = n, fast = n;
        do {
            slow = newNum(slow);
            fast = newNum(fast);
            fast = newNum(fast);
        } while (slow != fast);

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

        return false;
    }
}