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.

题解：
因为整数的范围非常大，而且有负数，所以用map来标记和查找元素。
无序找连续元素，可以从某一已有元素开始，递增，递减的查找和它连续的元素是否存在，找过的元素及时从map中删除。
本来想用hash_map，但是没有这个库。。。所以用了map
class  Solution {
 public :
     int  longestConsecutive(vector < int >   & num) {
         //  Start typing your C/C++ solution below
         //  DO NOT write int main() function
        map < int ,  int >  hmap;
        hmap.clear();
         int  n  =  num.size();
         for ( int  i = 0 ; i < n; i ++ ){
            hmap.insert(pair < int ,  int > (num[i], i));
        }
         int  ans = 0 , cnt = 0 ;
        map < int ,  int > ::iterator it;
         for ( int  i = 0 ; i < num.size(); i ++ ){
             int  cur  =  num[i];
            it  =  hmap.find(num[i]);
            cnt ++ ;
             if (it != hmap.end()){
                map < int ,  int > ::iterator iter;
                 while ( 1 ){
                    iter  =  hmap.find( ++ cur);
                     if (iter == hmap.end())
                         break ;
                    cnt ++ ;    
                    hmap.erase(iter);
                }
                cur = num[i];
                 while ( 1 ){
                    iter  =  hmap.find( -- cur);
                     if (iter == hmap.end())
                         break ;
                    cnt ++ ;    
                    hmap.erase(iter);
                }
                 if (ans < cnt)
                    ans  =  cnt;           
                cnt = 0 ;
            }
            cnt = 0 ;
        }
         return  ans;
    }
};