[LeetCode] 258. Add Digits

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

Example:

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

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

题意是给一个整数，求每一位上的数的加和。

两种做法，第一种是用递归，比较直观。对数字取余得到最低位的数字，然后再做加法直到num === 0。

JavaScript实现

/**
 * @param {number} num
 * @return {number}
 */
var addDigits = function(num) {
    let res = 0;
    while (num != 0) {
        res += num % 10;
        num = parseInt(num / 10);
    }
    if (res >= 10) {
        return addDigits(res);
    } else {
        return res;
    }
};

Java实现

class Solution {
    public int addDigits(int num) {
        int res = 0;
        while (num != 0) {
            res += num % 10;
            num /= 10;
        }
        if (res >= 10) {
            return addDigits(res);
        } else {
            return res;
        }
    }
}

第二种是看到了每次加和的规律，具体解释参见grandyang大神的第二个解法，https://www.cnblogs.com/grandyang/p/4741028.html

JavaScript实现

/**
 * @param {number} num
 * @return {number}
 */
var addDigits = function(num) {
    return ((num - 1) % 9) + 1;
};

Java实现

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

LeetCode 题目总结