3Sum
求和为指定值的三个数，麻烦的一点是，结果的集合不能有重复的。主要思路有两个，一个是在求值过程中过滤去重，还有一个是用hash。当然不能偷懒直接用set<vector<int> >，这样会直接超时。
// Dedup directly,
// LeetCode Judge Large, 272 milli secs.
vector<vector<int> > three_sum(vector<int> &num)
{
vector<vector<int> > ret;
if (num.size() == 0) return ret;
sort(num.begin(), num.end());
for (vector<int>::const_iterator it = num.begin();
it != num.end();
++it)
{
// Dedup
if (it != num.begin() && *it == *(it - 1))
{
continue;
}
// Dedup, front = it + 1
vector<int>::const_iterator front = it + 1;
vector<int>::const_iterator back = num.end() - 1;
while (front < back)
{
const int sum = *it + *front + *back;
if (sum > 0)
{
--back;
}
else if (sum < 0)
{
++front;
}
// Dedup
else if (front != it + 1 && *front == *(front - 1))
{
++front;
}
// Dedup
else if (back != num.end() - 1 && *back == *(back + 1))
{
--back;
}
else
{
vector<int> result;
// Already sorted.
result.push_back(*it);
result.push_back(*front);
result.push_back(*back);
ret.push_back(result);
++front;
--back;
}
}
}
return ret;
}