题目描述：

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.

最好想到的思路是排序，如果排序，最快也要O（NlogN），不符合要求。

以空间换时间：

代码：

int longestConsecutive(vector<int>& nums)

{

    map<int,int> map1;

    int len = 0;

    int maxlen = len;

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

    {

       //在nums中出现，则不为0，否则为0

       map1[nums[i]]++;

    }

    int min1 = nums[0];

    int min_up = min1;

    int min_down = min1-1;

    int i = 0;

    while(i<nums.size())

    {

          if(map1[min_up]!=0)

          {

                len++;

                map1[min_up] = 0;

                min_up++;

          }

          if(map1[min_down]!=0)

          {

                 len++;

                 map1[min_down] = 0;

                 min_down--;

          }

          if(map1[min_down]==0 && map1[min_up]==0)

          {

                 if(maxlen<len)

                 maxlen = len;

                 cout<<maxlen<<endl;

                 len = 0;

                 while(map1[nums[i++]]==0&&i<nums.size());

                 if(i<nums.size())

                 {

                      min1 = nums[i-1];

                      min_up = min1;

                      min_down = min1-1;

                 }

                 else break;

          }

    }

    return maxlen;

}

时间复杂度：O(N),同时引入了map，空间复杂度O(N).