Leetcode: 4Sum

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

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)

递归：judge small：Accepted! judge large: Time Limit Exceeded

void fourSumHelper(vector &num, int index, vector &path, int cursum, int target, vector> &res)
    {
    	if(path.size() >=4)
		{
			if(path.size() == 4 && cursum == target)
			{
				res.push_back(path);
				return;
			}else
				return;
		}
		for(int i = index; i < num.size(); i++)
    	{
			path.push_back(num[i]);
			cursum += num[i];
			fourSumHelper(num,i+1,path,cursum,target,res);
			path.pop_back();
			cursum -= num[i];
			while(i < num.size()-1 && num[i] == num[i+1])i++;
		}
	}
	vector > fourSum(vector &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector> res;
		if(num.size()<4)return res;
		sort(num.begin(), num.end());
		vector path;
		fourSumHelper(num,0,path,0,target,res);
		return res;
    }

2sum变形：Accepted!

vector > fourSum(vector &num, int target) {
        // Note: The Solution object is instantiated only once.
        vector> res;
    	int numlen = num.size();
		if(num.size()<4)return res;
		
		sort(num.begin(),num.end());
		set> tmpres;
		for(int i = 0; i < numlen; i++)
		{
			for(int j = i+1; j < numlen; j++)
			{
				int begin = j+1;
				int end = numlen-1;
				while(begin < end)
				{
					int sum = num[i]+ num[j] + num[begin] + num[end];
					if(sum == target)
					{
						vector tmp;
						tmp.push_back(num[i]);
						tmp.push_back(num[j]);
						tmp.push_back(num[begin]);
						tmp.push_back(num[end]);
						tmpres.insert(tmp);
						begin++;
						end--;
					}else if(sum>::iterator it = tmpres.begin();
		for(; it != tmpres.end(); it++)
			res.push_back(*it);
		return res;
    }