Longest Alternating Subarray
<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. A subarray <code>s</code> of length <code>m</code> is called <strong>alternating</strong> if:</p>

<ul>
	<li><code>m</code> is greater than <code>1</code>.</li>
	<li><code>s<sub>1</sub> = s<sub>0</sub> + 1</code>.</li>
	<li>The <strong>0-indexed</strong> subarray <code>s</code> looks like <code>[s<sub>0</sub>, s<sub>1</sub>, s<sub>0</sub>, s<sub>1</sub>,...,s<sub>(m-1) % 2</sub>]</code>. In other words, <code>s<sub>1</sub> - s<sub>0</sub> = 1</code>, <code>s<sub>2</sub> - s<sub>1</sub> = -1</code>, <code>s<sub>3</sub> - s<sub>2</sub> = 1</code>, <code>s<sub>4</sub> - s<sub>3</sub> = -1</code>, and so on up to <code>s[m - 1] - s[m - 2] = (-1)<sup>m</sup></code>.</li>
</ul>

<p>Return <em>the maximum length of all <strong>alternating</strong> subarrays present in </em><code>nums</code> <em>or </em><code>-1</code><em> if no such subarray exists</em><em>.</em></p>

<p>A subarray is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [2,3,4,3,4]</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

<p><strong>Explanation:</strong></p>

<p>The alternating subarrays are <code>[2, 3]</code>, <code>[3,4]</code>, <code>[3,4,3]</code>, and <code>[3,4,3,4]</code>. The longest of these is <code>[3,4,3,4]</code>, which is of length 4.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [4,5,6]</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p><code>[4,5]</code> and <code>[5,6]</code> are the only two alternating subarrays. They are both of length 2.</p>
</div>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>
Array; Enumeration
class Solution {
    public int alternatingSubarray(int[] nums) {
        int ans = -1, n = nums.length;
        for (int i = 0; i < n; ++i) {
            int k = 1;
            int j = i;
            for (; j + 1 < n && nums[j + 1] - nums[j] == k; ++j) {
                k *= -1;
            }
            if (j - i + 1 > 1) {
                ans = Math.max(ans, j - i + 1);
            }
        }
        return ans;
    }
}