LeetCode128：Longest Consecutive Sequence

题目：

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.

解题思路：

考虑到题目要求的时间复杂度O(n)，很自然的想到哈希表，只有哈希表的查找时间为O（1）。

实现代码：

#include <iostream>

#include <vector>

#include <unordered_set>

using namespace std;





/*

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.

*/

class Solution {

public:

    int longestConsecutive(vector<int> &num) {

        if(num.empty())

            return 0;

        unordered_set<int> iset;

        vector<int>::iterator iter;

        for(iter = num.begin(); iter != num.end(); ++iter)

            iset.insert(*iter);

        

        int max = 0;

        for(iter = num.begin(); iter != num.end(); ++iter)

        {

            if(iset.count(*iter) == 1)

            {

                int c = 1;

                int g = *iter + 1;

                while(iset.count(g) == 1)

                {

                    iset.erase(g);//找到后记得要删除，不然会重复做，超时 

                    c++;

                    g++;

                }

                int l = *iter - 1;

                while(iset.count(l) == 1)

                {

                    iset.erase(l);

                    c++;

                    l--;

                }

                if(max < c)

                    max = c;

                

            }

            

        }

        return max;    

    }

};

int main(void)

{

    int arr[] = {100, 4, 200, 1, 3, 2};

    int n = sizeof(arr) / sizeof(arr[0]);

    vector<int> num(arr, arr+n);

    Solution solution;

    int max = solution.longestConsecutive(num);

    cout<<max<<endl;

    return 0;

}