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)

 Solution:

public class Solution {
public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
	Arrays.sort(num);
 
	HashSet<ArrayList<Integer>> hashSet = new HashSet<ArrayList<Integer>>();
	ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
 
	for (int i = 0; i < num.length; i++) {
		for (int j = i + 1; j < num.length; j++) {
			int k = j + 1;
			int l = num.length - 1;
 
			while (k < l) {
				int sum = num[i] + num[j] + num[k] + num[l];
 
				if (sum > target) {
					l--;
				} else if (sum < target) {
					k++;
				} else if (sum == target) {
					ArrayList<Integer> temp = new ArrayList<Integer>();
					temp.add(num[i]);
					temp.add(num[j]);
					temp.add(num[k]);
					temp.add(num[l]);
 
					if (!hashSet.contains(temp)) {
						hashSet.add(temp);
						result.add(temp);
					}
 
					k++;
					l--;
				}
			}
		}
	}
 
	return result;
}

}

C++

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<int> tmp;
        vector<vector<int>> res;
        if(num.empty()) return res;
        sort(num.begin(), num.end());
        for(int i=0; i<num.size(); i++)
        {
            int cur = target - num[i];
            for(int j=i+1; j<num.size(); j++)
            {
                int temp = cur - num[j];
                int start = j+1, end = num.size()-1;
                while(start<end)
                {
                    if(num[start]+num[end]==temp)
                    {
                        tmp.push_back(num[i]);
                        tmp.push_back(num[j]);
                        tmp.push_back(num[start]);
                        tmp.push_back(num[end]);
                        res.push_back(tmp);
                        tmp.clear();
                        start++;
                        end--;
                        while(start<end&&num[start]==num[start-1]) start++;
                        while(start<end&&num[end]==num[end+1]) end--;
                    }
                    else if(num[start]+num[end]<temp)
                    {
                        start++;
                        while(start<end&&num[start]==num[start-1]) start++;
                    }
                    else
                    {
                        end--;
                        while(start<end&&num[end]==num[end+1]) end--;
                    }
                }
                while(j<num.size()&&num[j]==num[j+1]) j++; 
            }
            while(i<num.size()&&num[i]==num[i+1]) i++;
        }
        return res;
    }
};