k-sum问题

给定一个数组，里面的是任意整数，可能有重复，再给定一个目标T，从数组中找出所有和为T的K个数，要求结果中没有重复。

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.
 
    A solution set is:
    (-1,  0, 0, 1)
    (-2, -1, 1, 2)
    (-2,  0, 0, 2)

思路很容易就想到，还是利用将问题规模最小化的原则，简化问题成两个数的和为M的问题。整体的思想就是动态规划的思想了，只是特别需要注意的是要找出所有的可能。另外还需要特别处理重复和每个可能的内部顺序。4-sum的代码如下：

vector<vector<int>> fourSum(vector<int>& nums, int target) {
	std::sort(nums.begin(), nums.end());
 
	vector<vector<int>> ans;
	if (nums.size() >= 4)
	{
		findSum(ans, nums, target, 4, 0);
		std::set<vector<int> > dset;
		vector<vector<int> >::iterator itr = ans.begin();
		while (itr != ans.end())
		{
			std::reverse(itr->begin(), itr->end());
			dset.insert(*itr);
			++itr;
		}
		ans.clear();
		std::set<vector<int> >::iterator itr2 = dset.begin();
		while (itr2 != dset.end())
		{
			ans.push_back(*itr2++);
		}
	}
	return ans;
}
void findSum(vector<vector<int> > &ans, vector<int>& nums, int target, int count, int start) {
	int sz = nums.size();
	if (!(count < 2 || start > sz - 1))
	{
		int p = nums[start];
		if (count > 2)
		{
			findSum(ans, nums, target - p, count - 1, start + 1);
			vector<vector<int> >::iterator itr = ans.end();
			while (itr !=ans.begin())
			{
				--itr;
				if (itr->size() == count - 1)
				{
					itr->push_back(p);
				}
				else
					break;
			}
			findSum(ans, nums, target, count, start + 1);
		}
		else if (count == 2)
		{
			int i = start;
			int j = sz - 1;
			while (i < j)
			{
				if (target == nums[i] + nums[j])
				{
					vector<int> v;
					v.reserve(4);
					v.push_back(nums[j]);
					v.push_back(nums[i]);
					ans.push_back(v);
					i++;
				}
				else if ( nums[i] + nums[j] > target)
					j--;
				else
					i++;
			}
		}
	}
}