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;
}