Add Digits
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?
解題思路:
解法一:
模擬法,不斷將數字拆分然後相加,直到只有一位數。
class Solution {
public:
int addDigits(int num) {
while(num/10>0){
int sum = 0;
while(num!=0){
sum += num%10;
num = num/10;
}
num = sum;
}
return num;
}
};
解法二:
題目要求沒有遞歸,沒有循環,且時間複雜度爲O(1)。那就找規律。
~input: 0 1 2 3 4 ...
output: 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 ....
注意到,後面都是1-9不斷循環。
有兩個公式:
d(n) = num%9 == 0 ? (num==0? 0 : 9) : num%9
d(n) = 1 + (n-1) % 9
因此代碼如下:
class Solution {
public:
int addDigits(int num) {
return 1 + (num-1)%9;
}
};
或者:
class Solution {
public:
int addDigits(int num) {
return num%9 == 0? (num==0?0:9) : num%9;
}
};