(easy)LeetCode 258.Add Digits
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?
方法1:常规做法,循环,不符合题意。
public class Solution {
public int addDigits(int num) {
while(num>9){
int p=0;
while(num!=0){
p+=num%10;
num=num/10;
}
num=p;
}
return num;
}
}
方法2:找到规律,一行代码
代码如下:
public class Solution {
public int addDigits(int num) {
return (num-1)%9+1;
}
}
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