[Swift]LeetCode258. 各位相加 | Add Digits

发布时间：2020-12-14 05:11:13 所属栏目：百科来源：网络整理

导读：Given a non-negative integer? num ,repeatedly add all its digits until the result has only one digit. Example: Input: Output: 2 Explanation: The process is like:,. ? Since has only one digit,return it.383 + 8 = 111 + 1 = 22 Follow up: Coul

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

Example:

Input: 
Output: 2 
Explanation: The process is like:,. 
?            Since  has only one digit,return it.
383 + 8 = 111 + 1 = 22

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

给定一个非负整数?num，反复将各个位上的数字相加，直到结果为一位数。

示例:

输入: 
输出: 2 
解释: 各位相加的过程为：,。 由于? 是一位数，所以返回 2。
383 + 8 = 111 + 1 = 22

进阶:
你可以不使用循环或者递归，且在 O(1) 时间复杂度内解决这个问题吗？

20ms

1 class Solution {
2     func addDigits(_ num: Int) -> Int {
3         return (num - 1)%9 + 1 
4     }
5 }

24ms

 1 class Solution {
 2     func addDigits(_ num: Int) -> Int {
 3         var num = num
 4 
 5         while num >= 10 {
 6             var x = 0
 7             while num > 0 {
 8                 x += num % 10
 9                 num /= 10
10             }
11             num = x
12         }
13 
14         return num
15     }
16 }

24ms

 1 class Solution {
 2     
 3     func addDigits(_ num: Int) -> Int {
 4         guard num > 9 else { return num }
 5         
 6         var digits = [Int]()
 7         var number = num
 8         
 9         while true {
10             digits.removeAll()
11             
12             while number > 0 {
13                 digits.append(number % 10)
14                 number /= 10
15             }
16             
17             if digits.count > 1 {
18                 number = digits.reduce(0,+)
19             } else {
20                 number = digits.first ?? 0
21                 break
22             }
23         }
24         
25         return number
26     }
27 }

（编辑：李大同）

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