Leetcode | | 128, longest consecutive Sequence

Problem:

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.

Hide TagsArrayTest instructions: Finding the longest contiguous subsequence in a random ordinal group (no order required)

Thinking:

(1) requires time complexity of O (n), can only use Hashtable.

(2) Unordered_map bottom is realized with Hashtable, can take as Hashtable use

Code

Class Solution {Public:int longestconsecutive (vector<int> &num) {unordered_map<int,int> Hashta        ble            for (int i = 0;i < Num.size (); i++) {if (Hashtable.find (num[i])! = Hashtable.end ()) continue;            int minus_1 = num[i]-1;            int plus_1 = num[i] + 1;            Unordered_map<int,int>::iterator Minus_1iter,plus_1iter;            Minus_1iter = Hashtable.find (minus_1);            Plus_1iter = Hashtable.find (plus_1); if (Minus_1iter! = Hashtable.end () && plus_1iter! = Hashtable.end ()) {Hashtable[num[i]] = Hashtable                [Minus_1] + hashtable[plus_1] + 1;                Hashtable[num[i]-hashtable[minus_1]] = hashtable[num[i]];            Hashtable[num[i] + hashtable[plus_1]] = hashtable[num[i]]; } else if (Minus_1iter = = Hashtable.end () && Plus_1iter = = Hashtable.end ()) {Hashtable[n            Um[i]] = 1; } else if (minus_1iter! = Hashtable.end ()) {Hashtable[num[i]] = Hashtable[minus_1] + 1;            Hashtable[num[i]-hashtable[minus_1]] = hashtable[num[i]];                } else {Hashtable[num[i]] = Hashtable[plus_1] + 1;            Hashtable[num[i] + hashtable[plus_1]] = hashtable[num[i]];        }}//find the maxlen int ans = int_min; for (Unordered_map<int,int>::iterator iter = Hashtable.begin (); ITER! = Hashtable.end (); ++iter) {if (iter-&        Gt;second > ans) ans = iter->second;    } return ans; }};

Leetcode | | 128, longest consecutive Sequence