【ATT】Longest Consecutive Sequence

Q:

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.

A：使用Map来处理。对cluster来说，最重要的是上界，下界和长度。

对于a[i]，只需要检查a[i]-1，a[i]+1

映射上界下界到区间长度。merge 相邻的cluster时，只需要更新上界和下界对应的区间长度。

//1. The key factors about a cluster is: lowest, highest, and length.
//2. Map lowest and highest to length. To merge two neighbor clusters, only need to update it's new lowest and highest, with new length.
//3. For every a[i], checking its neighbor a[i]-1 and a[i]+1 is enough.    
int longestConsecutive(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        if(num.empty())
            return 0;
            
        int maxlen = 1;            //注意，初始化时1，而不是0
        
        map<int,int> hmap;
        
        for(int i=0;i<num.size();i++)
        {
            if(hmap.find(num[i])!=hmap.end())  //duplidate
                continue;
            hmap[num[i]] = 1;
            if(hmap.find(num[i]-1)!=hmap.end())
                maxlen = max(maxlen,merge(hmap,num[i]-1,num[i]));
            if(hmap.find(num[i]+1)!=hmap.end())
                maxlen = max(maxlen,merge(hmap,num[i],num[i]+1));
        }
        
        return maxlen;
        
    }
    
    int merge(map<int,int>& hmap,int left,int right)
    {
        int lower = left-hmap[left]+1;//新区间的下届
        int upper = right+hmap[right]-1; //新区间的上届
        
        int len = upper - lower +1;
        hmap[lower] = len;
        hmap[upper] = len;
        return len;
    }