Day 15 - LeetCode 258. Add Digits

LeetCode 258. Add Digits

題目

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

For example:
Given num = 38, 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?

翻譯

將一個數字每個位數相加，直到剩個位數為止。
範例：
num = 38，則 3+8 = 11，1+1 = 2, 2是個為數，回傳2。
進階：
不用迴圈，遞迴解這個問題

想法

轉陣列跑迴圈或是遞回加起來當小於10就可以return, 進階的有參考公式, 除9的餘數

詳細參閱

https://en.wikipedia.org/wiki/Digital_root

https://discuss.leetcode.com/topic/26300/elegant-3-line-recursion-c

---

Code

/**
 * @param {number} num
 * @return {number}
 */
var addDigits = function(num) {
    //代表只有個位數, 可直接跳出
    if (num < 10) {
        return num;
    }

    let str = String(num).split('');
    let len = str.length - 1;
    let i = 0 ;
    //ex : [1, 2, 3] => [1, 3, 6]
    var array_sum = str.map(function(val){
        i += parseInt(val);
        return i;
    });
    //取陣列最後一位
    return addDigits(array_sum[len]);
};

/**
 * @param {number} num
 * @return {number}
 */
var addDigits = function(num) {
    
    if (num > 9) {
        num = addDigits(parseInt(num / 10) + num % 10);
    }
    return num;
};

Run

Your input
0
234
152
1
99
9

Your answer
0
9
8
1
9
9

Expected answer
0
9
8
1
9
9

Runtime: 112 ms