题目

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

Example:

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

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

题意

将一个整数的各位数相加得到新整数，重复该步骤直到最终得到的是一个一位数。

思路

常规方法就是不断拆分相加直到得到最终结果。

O(1)方法：具体解析见LeetCode: Add Digits - 非负整数各位相加。

代码实现

Java

迭代

  1. class Solution {
  2.     public int addDigits(int num) {
  3.         while (num > 9) {
  4.             num = solve(num);
  5.         }
  7.         return num;
  8.     }
  10.     private int solve(int num) {
  11.         int res = 0;
  12.         while (num != 0) {
  13.             res += num % 10;
  14.             num /= 10;
  15.         }
  16.         return res;
  17.     }
  18. }

公式计算

  1. class Solution {
  2.     public int addDigits(int num) {
  3.         return (num - 1) % 9 + 1;
  4.     }
  5. }

JavaScript

迭代

  1. /**
  2.  * @param {number} num
  3.  * @return {number}
  4.  */
  5. var addDigits = function (num) {
  6.   while (num > 9) {
  7.     let tmp = 0
  8.     while (num) {
  9.       tmp += num % 10
  10.       num = Math.trunc(num / 10)
  11.     }
  12.     num = tmp
  13.   }
  14.   return num
  15. }

公式计算

  1. /**
  2.  * @param {number} num
  3.  * @return {number}
  4.  */
  5. var addDigits = function (num) {
  6.   return (num - 1) % 9 + 1
  7. }