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202. Happy Number

中文文档

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

Python3

class Solution:
    def isHappy(self, n: int) -> bool:
        def get_next(n):
            s = 0
            while n > 0:
                n, digit = divmod(n, 10)
                s += digit**2
            return s

        visited = set()
        while n != 1 and n not in visited:
            visited.add(n)
            n = get_next(n)
        return n == 1

Java

class Solution {
    public boolean isHappy(int n) {
        Set<Integer> visited = new HashSet<>();
        while (n != 1 && !visited.contains(n)) {
            visited.add(n);
            n = getNext(n);
        }
        return n == 1;
    }

    private int getNext(int n) {
        int s = 0;
        while (n > 0) {
            int digit = n % 10;
            s += digit * digit;
            n /= 10;
        }
        return s;
    }
}

C++

class Solution {
public:
    bool isHappy(int n) {
        auto getNext = [](int n) {
            int res = 0;
            while (n) {
                res += pow(n % 10, 2);
                n /= 10;
            }
            return res;
        };
        int slow = n;
        int fast = getNext(n);
        while (slow != fast) {
            slow = getNext(slow);
            fast = getNext(getNext(fast));
        }
        return slow == 1;
    }
};

TypeScript

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;
    };

    let slow = n;
    let fast = getNext(n);
    while (slow !== fast) {
        slow = getNext(slow);
        fast = getNext(getNext(fast));
    }
    return fast === 1;
}

Rust

impl Solution {
    pub fn is_happy(n: i32) -> bool {
        let get_next = |mut n: i32| {
            let mut res = 0;
            while n != 0 {
                res += (n % 10).pow(2);
                n /= 10;
            }
            res
        };
        let mut slow = n;
        let mut fast = get_next(n);
        while slow != fast {
            slow = get_next(slow);
            fast = get_next(get_next(fast));
        }
        slow == 1
    }
}

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