最长连续序列(O(n)，C++版)

/*given an unsorted array of integers, find the length of the longest consecutive elements sequence.

For example,
	Given [100, 4, 200, 1, 3, 2],
	The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.

	Your algorithm should run in O(n) complexity.*/

#include <vector>
#include <unordered_set>
#include <algorithm>
#include <iostream>

class Solution
{
public: 
	int longestConsecutive(std::vector<int>& num)
	{
		if(num.size()==0)
			return 0;
		std::unordered_set<int> set;
		int longest=0;
		int cc=0;
		std::for_each(num.begin(),num.end(),[&](const int& val)
				{
					set.insert(val);
				}
			);
		std::for_each(num.begin(),num.end(),[&](const int& val)
				{
					if(set.find(val)==set.end()) ;
					else
					{
						cc=1;
						set.erase(val);
						for(int tv=val+1;;++tv)
						{
							if(set.find(tv)!=set.end())
							{
								++cc;
								set.erase(tv);
							}
							else
								break;
						}
						for(int tv=val-1;;--tv)
						{
							if(set.find(tv)!=set.end())
							{
								++cc;
								set.erase(tv);
							}
							else
								break;
						}
						longest=std::max(cc,longest);
					}
				}
			);
		return longest;
	}
};

int main(int argc,char* argv[])
{
	int arr[]={100,4,200,1,3,2};

	std::vector<int> vec(arr,arr+sizeof(arr)/sizeof(int));
	Solution s;
	std::cout<<s.longestConsecutive(vec)<<std::endl;
	return 0;
}
/*given an unsorted array of integers, find the