3Sum

    求和為指定值的三個數，麻煩的一點是，結果的集合不能有重復的。主要思路有兩個，一個是在求值過程中過濾去重，還有一個是用hash。當然不能偷懶直接用set<vector<int> >，這樣會直接超時。

// 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;
}