258.Add Digits
Given an integer num, repeatedly add all its digits until the result has only one digit, and return it.
Example 1:
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| Input: num = 38
Output: 2
Explanation: The process is
38 --> 3 + 8 --> 11
11 --> 1 + 1 --> 2
Since 2 has only one digit, return it.
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Example 2:
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| Input: num = 0
Output: 0
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个人解答：
时间复杂度：O(log num)
空间复杂度：O(1)
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| class Solution {
public:
int addDigits(int num) {
int sum = 0;
do {
while(num != 0) {
sum += num%10;
num/=10;
}
num = sum;
sum = 0;
}while(num > 9);
return num;
}
};
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进阶：
时间复杂度：O(1)
空间复杂度：O(1)
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| class Solution {
public:
int addDigits(int num) {
return (num - 1) % 9 + 1;
}
};
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