0258. Add Digits (E)

Add Digits (E)

题目

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?

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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。

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代码实现

Java

迭代

class Solution {
    public int addDigits(int num) {
        while (num > 9) {
            num = solve(num);
        }

        return num;
    }

    private int solve(int num) {
        int res = 0;
        while (num != 0) {
            res += num % 10;
            num /= 10;
        }
        return res;
    }
}

公式计算

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

JavaScript

迭代

/**
 * @param {number} num
 * @return {number}
 */
var addDigits = function (num) {
  while (num > 9) {
    let tmp = 0
    while (num) {
      tmp += num % 10
      num = Math.trunc(num / 10)
    }
    num = tmp
  }
  return num
}

公式计算

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