main.cpp

// Source: https://leetcode.com/problems/self-dividing-numbers
// Title: Self Dividing Numbers
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// 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`.
//
// A **self-dividing number** is not allowed to contain the digit zero.
//
// Given two integers `left` and `right`, return a list of all the **self-dividing numbers** in the range `[left, right]` (both **inclusive**).
//
// **Example 1:**
//
// ```
// Input: left = 1, right = 22
// Output: [1,2,3,4,5,6,7,8,9,11,12,15,22]
// ```
//
// **Example 2:**
//
// ```
// Input: left = 47, right = 85
// Output: [48,55,66,77]
// ```
//
// **Constraints:**
//
// - `1 <= left <= right <= 10^4`
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

#include <vector>

using namespace std;

class Solution {
  static bool selfDivisible(int n) {
    int tmp = n;
    while (tmp > 0) {
      int digit = tmp % 10;
      if (digit == 0 || n % digit != 0) return false;
      tmp /= 10;
    }
    return true;
  }

 public:
  vector<int> selfDividingNumbers(int left, int right) {
    auto ans = vector<int>();
    ans.reserve(right - left + 1);

    for (int num = left; num <= right; ++num) {
      if (selfDivisible(num)) ans.push_back(num);
    }

    return ans;
  }
};