README_EN.md

258. Add Digits

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Description

Given an integer num, repeatedly add all its digits until the result has only one digit, and return it.

 

Example 1:

Input: num = 38
Output: 2
Explanation: The process is
38 --> 3 + 8 --> 11
11 --> 1 + 1 --> 2 
Since 2 has only one digit, return it.

Example 2:

Input: num = 0
Output: 0

 

Constraints:

• 0 <= num <= 231 - 1

 

Follow up: Could you do it without any loop/recursion in O(1) runtime?

Solutions

Solution 1

class Solution:
    def addDigits(self, num: int) -> int:
        return 0 if num == 0 else (num - 1) % 9 + 1

class Solution {
    public int addDigits(int num) {
        return (num - 1) % 9 + 1;
    }
}

class Solution {
public:
    int addDigits(int num) {
        return (num - 1) % 9 + 1;
    }
};

func addDigits(num int) int {
	if num == 0 {
		return 0
	}
	return (num-1)%9 + 1
}

impl Solution {
    pub fn add_digits(num: i32) -> i32 {
        if num < 10 {
            return num;
        }
        Self::add_digits(
            num
                .to_string()
                .chars()
                .map(|c| c.to_string().parse::<i32>().unwrap())
                .sum::<i32>()
        )
    }
}

Solution 2

impl Solution {
    pub fn add_digits(mut num: i32) -> i32 {
        ((num - 1) % 9) + 1
    }
}