leetcode笔记--6 Add Digits

question:

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.

answer:(which don‘t think by myself)

思路：

举例：

38,11,2    38-2=36（38-11=27,11-2=9，都是9的倍数）

57,12,3    57-3=54（57-12=45,12-3=9，都是9的倍数）

所以直接用57/9 得余数为3 ？？

其实不对.......忘记了特殊情况（就是整除的时候）

18,9

余数为0，不符合上述方法

所以需要先减去一个数，最后余数再加上一个数，可以避免这种情况，但是可以随意先加后减一个数吗？？

不可以，举例10-2=8 8对9取余数为8,8+2=10，不符合条件

所以只能是1

时间： 2024-10-12 16:59:57

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