看眼缘随机拿了一道不难的题https://leetcode.com/problems/add-digits/description/
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?
提交测试通过:
int addDigits(int num) {
if(num < 10) return num;
int add = 0;
while(num > 0){
add += num % 10;
num = num/10;
}
return addDigits(add);
}
用了递归,用循环做也行
本文解析了LeetCode上一道关于数字操作的问题:如何通过反复累加非负整数的所有位直到只剩一位数字。提供了递归实现的方法,并讨论了是否能在O(1)时间内不使用循环或递归解决该问题。
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