[LeetCode] Self Dividing Numbers
阿新 • • 發佈:2017-11-20
i++ nta left vector because span || digi eve
A self-dividing number is a number that is divisible by every digit it contains.
For example, 128 is a self-dividing number because 128 % 1 == 0
, 128 % 2 == 0
, and 128 % 8 == 0
.
Also, a self-dividing number is not allowed to contain the digit zero.
Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.
Example 1:
Input: left = 1, right = 22 Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]
Note:
- The boundaries of each input argument are
1 <= left <= right <= 10000
.
找出給定區間內的自分數,自分數就是可以整除的數字中每一位數的那個數。由此判斷自分數中不含有數字0。
首先寫一個函數來判斷自分數,按照自分數的定義書寫即可。註意判斷數字0不能是除數。
然後在主函數中,循環添加符合要求的自分數。
class Solution {public: vector<int> selfDividingNumbers(int left, int right) { vector<int> res; for (int i = left; i <= right; i++) { if (isSelfDividingNumbers(i)) res.push_back(i); } return res; } bool isSelfDividingNumbers(int num) {int tmp = num, digit = 0; while (tmp) { digit = tmp % 10; if (digit == 0 || num % digit != 0) return false; tmp /= 10; } return true; } }; // 3 ms
[LeetCode] Self Dividing Numbers