题目:
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?
Show Hint
题目分析:
首先觉得例子很明显,可能的结果是数目固定的(0~9),那么很有可能是由规律的,加上后面的一句,要求在常数范围内做出来,那么很可能有很大的规律,从0~30总结 一下规律
总结得出结论是从1~9循环
我在审题时候,遗漏了非负数,包括0,没有考虑0的特殊情况
代码:
public class Solution {
public int addDigits(int num) {
if(num == 0){
return 0;
}else{
if(num%9 != 0){
return (num%9);
}else{
return 9;
}
}
}
}