<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;
        }
        sort(nums.begin(), nums.end());
        vector<vector<int>> seqs;
        vector<int> cur_seq;
        cur_seq.push_back(nums[0]);
        for (int i = 0; i < nums.size() - 1; i++) {
            if (nums[i] + 1 == nums[i + 1]) {
                cur_seq.push_back(nums[i + 1]);
            } else if (nums[i] == nums[i + 1]) {
                continue;
            } else {
                seqs.push_back(cur_seq);
                cur_seq.erase(cur_seq.begin(), cur_seq.end());
                cur_seq.push_back(nums[i + 1]);
            }
        }
        seqs.push_back(cur_seq);
        int longest = 1;
        for (int i = 0; i < seqs.size(); i++) {
            if (seqs[i].size() > longest) {
                longest = seqs[i].size();
            }
        }
        return longest;
    }
};
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