Smallest Value of the Rearranged Number
<p>You are given an integer <code>num.</code> <strong>Rearrange</strong> the digits of <code>num</code> such that its value is <strong>minimized</strong> and it does not contain <strong>any</strong> leading zeros.</p>

<p>Return <em>the rearranged number with minimal value</em>.</p>

<p>Note that the sign of the number does not change after rearranging the digits.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> num = 310
<strong>Output:</strong> 103
<strong>Explanation:</strong> The possible arrangements for the digits of 310 are 013, 031, 103, 130, 301, 310. 
The arrangement with the smallest value that does not contain any leading zeros is 103.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> num = -7605
<strong>Output:</strong> -7650
<strong>Explanation:</strong> Some possible arrangements for the digits of -7605 are -7650, -6705, -5076, -0567.
The arrangement with the smallest value that does not contain any leading zeros is -7650.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>-10<sup>15</sup> &lt;= num &lt;= 10<sup>15</sup></code></li>
</ul>
Math; Sorting
class Solution {
public:
    long long smallestNumber(long long num) {
        bool neg = num < 0;
        num = abs(num);
        int cnt[10]{};
        while (num > 0) {
            ++cnt[num % 10];
            num /= 10;
        }
        long long ans = 0;
        if (neg) {
            for (int i = 9; i >= 0; --i) {
                while (cnt[i] > 0) {
                    ans = ans * 10 + i;
                    --cnt[i];
                }
            }
            return -ans;
        }
        if (cnt[0]) {
            for (int i = 1; i < 10; ++i) {
                if (cnt[i] > 0) {
                    --cnt[i];
                    ans = i;
                    break;
                }
            }
        }
        for (int i = 0; i < 10; ++i) {
            while (cnt[i] > 0) {
                ans = ans * 10 + i;
                --cnt[i];
            }
        }
        return ans;
    }
};
