LeetCode Add Digits
2016-03-01 03:39
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Given a non-negative integer
one digit.
For example:
Given
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 method?
这道题让我们求数根,所谓树根,就是将大于10的数的各个位上的数字相加,若结果还大于0的话,则继续相加,直到数字小于10为止。那么根据这个性质,我们可以写出一个解法如下:
解法一:
但是这个解法在出题人看来又trivial又naive,需要想点高逼格的解法,一行搞定碉堡了,那么我们先来观察1到20的所有的树根:
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 1
11 2
12 3
13 4
14 5
15 6
16 7
17 8
18 9
19 1
20 2
根据上面的列举,我们可以得出规律,每9个一循环,所有大于9的数的树根都是对9取余,那么对于等于9的数对9取余就是0了,为了得到其本身,而且同样也要对大于9的数适用,我们就用(n-1)%9+1这个表达式来包括所有的情况,所以解法如下:
解法二:
转自http://www.cnblogs.com/grandyang/p/4741028.html
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
2has
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 method?
这道题让我们求数根,所谓树根,就是将大于10的数的各个位上的数字相加,若结果还大于0的话,则继续相加,直到数字小于10为止。那么根据这个性质,我们可以写出一个解法如下:
解法一:
class Solution { public: int addDigits(int num) { while (num / 10 > 0) { int sum = 0; while (num > 0) { sum += num % 10; num /= 10; } num = sum; } return num; } };
但是这个解法在出题人看来又trivial又naive,需要想点高逼格的解法,一行搞定碉堡了,那么我们先来观察1到20的所有的树根:
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 1
11 2
12 3
13 4
14 5
15 6
16 7
17 8
18 9
19 1
20 2
根据上面的列举,我们可以得出规律,每9个一循环,所有大于9的数的树根都是对9取余,那么对于等于9的数对9取余就是0了,为了得到其本身,而且同样也要对大于9的数适用,我们就用(n-1)%9+1这个表达式来包括所有的情况,所以解法如下:
解法二:
class Solution { public: int addDigits(int num) { return (num - 1) % 9 + 1; } };
转自http://www.cnblogs.com/grandyang/p/4741028.html
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