[leetcode-128-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：

建立一个hash表，查找是否在表里面，向前和向后查找，比如对于数组元素100，然后查找是否存在99,98...等，还有是否

存在101,102...等。表里不存在就继续查找数组中别的元素。

比如，对于4，往左查找到3,2,1等，最长序列长度就为4。

还有就是，如果已经查到某元素，然后就将其在表里删除，能够减少后续查找次数。

int longestConsecutive(vector<int>& nums)
    {
        if (nums.size() == 0)return 0;
        unordered_set<int>record(nums.begin(),nums.end());//hash表格
        int result = 1;
        for (int i = 0; i < nums.size();i++)
        {
            if (record.find(nums[i]) == record.end()) continue;//没有找到i
            record.erase(nums[i]);//如果在的话就删除i，减少set的规模
            
            int pre = nums[i] - 1, next = nums[i] + 1;//向前和向后查找
            while (record.find(pre) != record.end())
            {
                record.erase(pre);
                pre--;
            }
            while (record.find(next) != record.end())
            {
                record.erase(next);
                next++;
            }

            result = max(result,next-pre-1);
        }
        return result;
    }

参考：

https://discuss.leetcode.com/topic/16483/a-simple-c-solution-using-unordered_set-and-simple-consideration-about-this-problem