Array Nesting
<p>You are given an integer array <code>nums</code> of length <code>n</code> where <code>nums</code> is a permutation of the numbers in the range <code>[0, n - 1]</code>.</p>

<p>You should build a set <code>s[k] = {nums[k], nums[nums[k]], nums[nums[nums[k]]], ... }</code> subjected to the following rule:</p>

<ul>
	<li>The first element in <code>s[k]</code> starts with the selection of the element <code>nums[k]</code> of <code>index = k</code>.</li>
	<li>The next element in <code>s[k]</code> should be <code>nums[nums[k]]</code>, and then <code>nums[nums[nums[k]]]</code>, and so on.</li>
	<li>We stop adding right before a duplicate element occurs in <code>s[k]</code>.</li>
</ul>

<p>Return <em>the longest length of a set</em> <code>s[k]</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [5,4,0,3,1,6,2]
<strong>Output:</strong> 4
<strong>Explanation:</strong> 
nums[0] = 5, nums[1] = 4, nums[2] = 0, nums[3] = 3, nums[4] = 1, nums[5] = 6, nums[6] = 2.
One of the longest sets s[k]:
s[0] = {nums[0], nums[5], nums[6], nums[2]} = {5, 6, 2, 0}
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,1,2]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt; nums.length</code></li>
	<li>All the values of <code>nums</code> are <strong>unique</strong>.</li>
</ul>
Depth-First Search; Array
class Solution {
public:
    int arrayNesting(vector<int>& nums) {
        int n = nums.size();
        vector<bool> vis(n);
        int res = 0;
        for (int i = 0; i < n; ++i) {
            if (vis[i]) continue;
            int cur = nums[i], m = 1;
            vis[cur] = true;
            while (nums[cur] != nums[i]) {
                cur = nums[cur];
                ++m;
                vis[cur] = true;
            }
            res = max(res, m);
        }
        return res;
    }
};