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.

思路：这道题求出最长连续数组，先从小到达排序，然后删除重复元素，在进行循环比较；如果相邻的元素相差为1，这maxlen加1，如果不是则maxlen=1，result[maxlen]=A[i]重新来过。但是这道题要求时间复杂度是O(n)，用到排序了时间复杂度变为O(nlogn).

class Solution {

public:

    int longestConsecutive(vector<int> &num) {

        int n=num.size();

        if(n<=1)

            return n;

        sort(num.begin(),num.end());

        num.erase(unique(num.begin(),num.end()),num.end());

        int maxlen=1;

        int result[n];

        result[maxlen]=num[0];

        int length=1;

        for(int i=1;i<num.size();i++)

        {

            if(num[i]-num[i-1]==1)

            {

                result[maxlen++]=num[i];

            }

            else

            {

                maxlen=1;

                result[maxlen]=num[i];

            }

            if(maxlen>length)

                length=maxlen;

        }

        return length;

    }

};

解法二：时间复杂度为O(n)，从当前元素递减寻找，从当前元素+1递增寻找，这样一次就能够把连续序列找全。找全后，别忘记清空元素，保证下次不会重复寻找，以减少重复搜寻的次数。

class Solution {

public:

    int longestConsecutive(vector<int> &num) {

        map<int,int> result;

        for(int i=0;i<num.size();i++)

        {

            result[num[i]]=1;

        }

        int maxlen=0;

        for(int i=0;i<num.size();i++)

        {

            int sum=1;

            if(result.count(num[i]))

            {

                result[num[i]]=0;

                int left=num[i]-1;

                while(result.count(left)&&result[left]!=0)

                {

                    result[left]=0;

                    left--;

                    sum++;

                }

                int right=num[i]+1;

                while(result.count(right)&&result[right]!=0)

                {

                    result[right]=0;

                    right++;

                    sum++;

                }

            }

            if(maxlen<sum)

                maxlen=sum;

        }

        return maxlen;

    }

};