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?
Hint:
- A naive implementation of the above process is trivial. Could you come up with other methods?
- What are all the possible results?
- How do they occur, periodically or randomly?
- You may find this Wikipedia article useful.
解题思路
思路1:一直相加直到num变为个位数。
思路2:套用维基公式。
实现代码
Java代码 1:
// Runtime: 2 ms
public class Solution {
public int addDigits(int num) {
int temp = 0;
while (num > 9) {
while (num != 0) {
temp += num % 10;
num /= 10;
}
num = temp;
temp = 0;
}
return num;
}
}
Java代码2:
// Runtime: 5 ms
public class Solution {
public int addDigits(int num) {
return (int)(num - 9 * Math.ceil((num - 1) / 9));
}
}
Java代码3:
// Runtime: 2 ms
public class Solution {
public int addDigits(int num) {
return 1 + (num - 1) % 9;
}
}
时间: 2024-09-26 09:55:53