202. Happy Number

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.

 

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

Solutions

Solution 1

Python3JavaC++GoTypeScriptRustC

class Solution:
    def isHappy(self, n: int) -> bool:
        vis = set()
        while n != 1 and n not in vis:
            vis.add(n)
            x = 0
            while n:
                n, v = divmod(n, 10)
                x += v * v
            n = x
        return n == 1

class Solution {
    public boolean isHappy(int n) {
        Set<Integer> vis = new HashSet<>();
        while (n != 1 && !vis.contains(n)) {
            vis.add(n);
            int x = 0;
            while (n != 0) {
                x += (n % 10) * (n % 10);
                n /= 10;
            }
            n = x;
        }
        return n == 1;
    }
}

class Solution {
public:
    bool isHappy(int n) {
        unordered_set<int> vis;
        while (n != 1 && !vis.count(n)) {
            vis.insert(n);
            int x = 0;
            for (; n; n /= 10) {
                x += (n % 10) * (n % 10);
            }
            n = x;
        }
        return n == 1;
    }
};

func isHappy(n int) bool {
    vis := map[int]bool{}
    for n != 1 && !vis[n] {
        vis[n] = true
        x := 0
        for ; n > 0; n /= 10 {
            x += (n % 10) * (n % 10)
        }
        n = x
    }
    return n == 1
}

function isHappy(n: number): boolean {
    const getNext = (n: number) => {
        let res = 0;
        while (n !== 0) {
            res += (n % 10) ** 2;
            n = Math.floor(n / 10);
        }
        return res;
    };
    const set = new Set();
    while (n !== 1) {
        const next = getNext(n);
        if (set.has(next)) {
            return false;
        }
        set.add(next);
        n = next;
    }
    return true;
}

use std::collections::HashSet;
impl Solution {
    fn get_next(mut n: i32) -> i32 {
        let mut res = 0;