<p>Given an unsorted array of integers <code>nums</code>, return <em>the length of the longest consecutive elements sequence.</em></p>

<p>You must write an algorithm that runs in&nbsp;<code>O(n)</code>&nbsp;time.</p>

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

<pre>
<strong>Input:</strong> nums = [100,4,200,1,3,2]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest consecutive elements sequence is <code>[1, 2, 3, 4]</code>. Therefore its length is 4.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,3,7,2,5,8,4,6,0,1]
<strong>Output:</strong> 9
</pre>

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

<ul>
	<li><code>0 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
  int longestConsecutive(vector<int> &nums) {
    if (nums.empty())
      return 0;

    int ans = 1;
    unordered_set<int> numSet;
    numSet.reserve(2 * nums.size());

    for (auto &num : nums) {
      numSet.insert(num + 1e5L);
    }

    while (!numSet.empty()) {
      // auto iter = numSet.begin();

      int l = *numSet.begin();
      int r = l;

      numSet.erase(l);

      // search left
      while (numSet.contains(l - 1)) {
        numSet.erase(l);
        l = l - 1;
      }

      // search right
      while (numSet.contains(r + 1)) {
        numSet.erase(r);
        r = r + 1;
      }

      ans = max(ans, r - l + 1);
    }

    return ans;
  }
};
71604