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;
}