4Sum

    求数组中和为target的4个数，结果的4个数是非降序，结果集合不能有重复。跟3Sum一样的解法。

vector<vector<int> > fourSum(vector<int> &num, int target)
{
    vector<vector<int> > ret;

    if (num.size() == 0) return ret;

    sort(num.begin(), num.end());

    for (size_t a = 0; a < num.size(); ++a)
    {
        if (a != 0 && num[a] == num[a - 1])
        {
            continue;
        }

        for (size_t b = a + 1; b < num.size(); ++b)
        {
            if (b != a + 1 && num[b] == num[b - 1])
            {
                continue;
            }

            size_t c = b + 1;
            size_t d = num.size() - 1;

            while (c < d)
            {
                const int sum = num[a] + num[b] + num[c] + num[d];

                if (sum > target)
                {
                    --d;
                }
                else if (sum < target)
                {
                    ++c;
                }
                else if (c != b + 1 && num[c] == num[c - 1])
                {
                    ++c;
                }
                else if (d != num.size() - 1 && num[d] == num[d + 1])
                {
                    --d;
                }
                else
                {
                    vector<int> result;

                    result.push_back(num[a]);
                    result.push_back(num[b]);
                    result.push_back(num[c]);
                    result.push_back(num[d]);

                    ret.push_back(result);

                    ++c;
                    --d;
                }
            }
        }
    }

    return ret;
}