Minimum Factorization
<p>Given a positive integer num, return <em>the smallest positive integer </em><code>x</code><em> whose multiplication of each digit equals </em><code>num</code>. If there is no answer or the answer is not fit in <strong>32-bit</strong> signed integer, return <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> num = 48
<strong>Output:</strong> 68
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> num = 15
<strong>Output:</strong> 35
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= num &lt;= 2<sup>31</sup> - 1</code></li>
</ul>
Greedy; Math
class Solution {
    public int smallestFactorization(int num) {
        if (num < 2) {
            return num;
        }
        long ans = 0, mul = 1;
        for (int i = 9; i >= 2; --i) {
            if (num % i == 0) {
                while (num % i == 0) {
                    num /= i;
                    ans = mul * i + ans;
                    mul *= 10;
                }
            }
        }
        return num < 2 && ans <= Integer.MAX_VALUE ? (int) ans : 0;
    }
}