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：

用unordered_map来记录数字是否在数组中。 然后枚举每个数，看其相邻的数在不在unorded_map中，相邻的数在数组里的话，每个数之多访问一次；相邻的数不在数组里的话，枚举会中断。所以设哈希复杂度为O(1)的话，这个方法是严格的O(n)。

代码：

class Solution {public:    int longestConsecutive(vector
     
       &num) {        unordered_map
      
        dict;        //init        for (int i = 0; i < num.size(); i++) {            dict[num[i]] = false;        }        int ans = 0;        for (int i = 0; i < num.size(); i++) {            if (dict[num[i]] == false) {                int cnt = 0;                //枚举相邻的数，看是否在字典中                for (int j = num[i]; dict.count(j); j++, cnt++) {                    dict[j] = true;                }                for (int j = num[i] - 1; dict.count(j); j--, cnt++) {                    dict[j] = true;                }                ans = max(ans, cnt);            }        }        return ans;    }};