LeetCode "Longest Consecutive Sequence"

Comparison-based sorting takes O(nlgn), so hashset is a good idea. After keeping records of all numbers in the hashset, you start checking each number. If it is isolated (no n-1 nor n+1) that number will be removed from hashset; If not isolated, we do a BFS scanning. Since each number is only checked once, it is O(n) with 2 passes.

class Solution {
public:
    int longestConsecutive(vector<int> &num) {
        unordered_set<int> rec;
         
        for (int i : num )         
            if(rec.find(i) == rec.end())    rec.insert(i);
        
        int cnt = 1, currcnt = 0;
        while(!rec.empty())
        {
            int i = *rec.begin();
            if( (rec.find(i - 1) == rec.end()) && 
                (rec.find(i + 1) == rec.end()) )
            {
                rec.erase(i);
                continue;
            }

            queue<int> q; q.push(i);
            while(!q.empty())
            {
                int n = q.front();
                currcnt ++;
                q.pop(); rec.erase(n);
                if(rec.find(n - 1) != rec.end()) q.push(n - 1);
                if(rec.find(n + 1) != rec.end()) q.push(n + 1);
            }
            cnt = std::max(currcnt, cnt);
            currcnt = 0;            
        }
        return cnt;
    }
};