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.
First of all,if there is no time complexity,we could sort the array,and then scan the array.Unfortunately,the time complexity is O(n*log n).It‘s been many times,when I am asked to solve a problem in O(n) which can be solved in O(n*log n) by using classical quicksort I would use a alternative method--hash table to do it.Here is the code.
1 class Solution { 2 public: 3 int longestConsecutive(vector<int> &num) { 4 unordered_map<int,bool> hashmap; 5 int maxlength = 0; 6 int curlength = 0; 7 8 for(vector<int>::iterator iter = num.begin(); 9 iter!=num.end();++iter){ 10 hashmap.insert(pair<int,bool>(*iter,true)); 11 } 12 13 for(vector<int>::iterator iter = num.begin(); 14 iter!=num.end();++iter){ 15 int num = *iter; 16 curlength = 0; 17 while(hashmap.find(num) != hashmap.end()){ 18 ++curlength; 19 hashmap.erase(num); 20 ++num; 21 } 22 23 num = *iter-1; 24 while(hashmap.find(num) != hashmap.end()){ 25 ++curlength; 26 hashmap.erase(num); 27 --num; 28 } 29 30 maxlength = maxlength<curlength?curlength:maxlength; 31 } 32 return maxlength; 33 } 34 };
[LeetCode] Longest Consecutive Sequence
原文：http://www.cnblogs.com/zhouyoulie/p/4008824.html