Leetcode : 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：
题目咋一看很简单，排个序然后从头到尾遍历一下就好，但是这种方法并不能解决该问题，关键是该题要求算法的复杂度是O(n)，而排序最快也需要O(n*log(n))。那么只能用空间换时间了，所以很自然就想到了Hash。（之前自以为对Hash学的很好，直到用的时候才发现其实Hash还不存在我的思维中）。

下面是我的代码，其中用到了std::unordered_map（Unordered associative containers implement unsorted (hashed) data structures that can be quickly searched (O(1)amortized, O(n) worst-case complexity），它是C++11标准引入的东西，具体用法可参考http://en.cppreference.com/w/cpp/container/unordered_map

int longestConsecutive(vector<int> &num)
{
    if (num.empty()) return 0;

    if (1 == num.size()) return 1;

    unordered_map<int, int> num_map;

    for (vector<int>::iterator it = num.begin(); it != num.end(); ++it)
    {
        ++num_map[*it];
    }

    int maxLen = 1;

    for (vector<int>::iterator it = num.begin(); it!= num.end(); ++it)
    {
        if (num_map[*it] == 0)
            continue;

        int len = 1;
        num_map[*it] = 0;
        int low = *it - 1;
        int high = *it + 1;

        bool bFindLow = true;
        bool bFindHigh = true;


        while (true)
        {
            bool bFind = false;

            unordered_map<int, int>::iterator iterMap;
            iterMap = num_map.find(low);
            if (bFindLow && iterMap != num_map.end() && iterMap->second)
            {
                ++len;
                --low;
                iterMap->second = 0;
                bFind = true;   
            }
            else
                bFindLow = false;

            iterMap = num_map.find(high);
            if (bFindHigh && iterMap != num_map.end() && iterMap->second)
            {
                ++len;
                ++high;
                iterMap->second = 0;
                bFind = true;
            }
            else
                bFindHigh = false;

            if (!bFind)
                break;
        }

        if (len > maxLen)
            maxLen = len;
    }

    return maxLen;
}