128. Longest Consecutive Sequence

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.

The Union Find way:

We could use a hashmap to apply union find, once comes in a new number i , we connect i with (i - 1) and (i + 1) if they already exist in the union find. And we should always ensure that the root is the smallest number in the consecutive sequence.

Code:

public class Solution {    public int longestConsecutive(int[] nums) {        HashMap<Integer,Integer> uf = new HashMap();        for(int i : nums){            if(!uf.containsKey(i)){                uf.put(i,i);                if(uf.containsKey(i + 1)){                    union(uf,i,i+1);                }                if(uf.containsKey(i - 1)){                    union(uf,i-1,i);                }            }        }        int len = 0;        for(int key : new HashSet<Integer>(uf.keySet())){            int pre = findRoot(uf,key);            len = Math.max((key - pre + 1), len);        }        return len;            }        int findRoot(HashMap<Integer,Integer> uf, int x){        if(uf.containsKey(x)){            if(uf.get(x) != x){                int ans = findRoot(uf,uf.get(x));                uf.put(x,ans);                return ans;            }        }        return x;    }        void union(HashMap<Integer,Integer> uf, int x, int y){        int r1 = findRoot(uf,x);        int r2 = findRoot(uf,y);        if(r1 > r2){            uf.put(r1,r2);        } else {            uf.put(r2,r1);        }    }}

Another way to do it is to convert the array to a set, for each number i , if i is the start of the sequence( (i - 1) not in the set), count the length of the sequence starting at i, otherwise continue.

Code:

public class Solution {    public int longestConsecutive(int[] nums) {        HashSet<Integer> hs = new HashSet();        for(int i : nums){            hs.add(i);        }        int ret = 0;        for(int i : hs){            if(!hs.contains(i - 1)){                int c = 0;                while(hs.contains(i++)){                    c++;                }                ret = Math.max(ret,c);            }                    }        return ret;            }}