Count Symmetric Integers
<p>You are given two positive integers <code>low</code> and <code>high</code>.</p>

<p>An integer <code>x</code> consisting of <code>2 * n</code> digits is <strong>symmetric</strong> if the sum of the first <code>n</code> digits of <code>x</code> is equal to the sum of the last <code>n</code> digits of <code>x</code>. Numbers with an odd number of digits are never symmetric.</p>

<p>Return <em>the <strong>number of symmetric</strong> integers in the range</em> <code>[low, high]</code>.</p>

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

<pre>
<strong>Input:</strong> low = 1, high = 100
<strong>Output:</strong> 9
<strong>Explanation:</strong> There are 9 symmetric integers between 1 and 100: 11, 22, 33, 44, 55, 66, 77, 88, and 99.
</pre>

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

<pre>
<strong>Input:</strong> low = 1200, high = 1230
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are 4 symmetric integers between 1200 and 1230: 1203, 1212, 1221, and 1230.
</pre>

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

<ul>
	<li><code>1 &lt;= low &lt;= high &lt;= 10<sup>4</sup></code></li>
</ul>
Math; Enumeration
public class Solution {
    public int CountSymmetricIntegers(int low, int high) {
        int ans = 0;
        for (int x = low; x <= high; ++x) {
            ans += f(x);
        }
        return ans;
    }

    private int f(int x) {
        string s = x.ToString();
        int n = s.Length;
        if (n % 2 == 1) {
            return 0;
        }
        int a = 0, b = 0;
        for (int i = 0; i < n / 2; ++i) {
            a += s[i] - '0';
        }
        for (int i = n / 2; i < n; ++i) {
            b += s[i] - '0';
        }
        return a == b ? 1 : 0;
    }
}
