4 Sum

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)

public class Solution {
    public ArrayList> fourSum(int[] num, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
        Arrays.sort(num);
        ArrayList<Integer> list = new ArrayList<Integer>();
        
        for (int i = 0; i < num.length - 3; i++) {
            int a = num[i];
            for (int j = i + 1; j < num.length - 2; j++) {
                int b = num[j];
                int k = j + 1;
                int l = num.length - 1;
                while (k < l) {
                    int c = num[k];
                    int d = num[l];
                    if (a + b + c + d == target) {
                        ArrayList sublist = new ArrayList();
                        sublist.add(a);
                        sublist.add(b);
                        sublist.add(c);
                        sublist.add(d);
                        list.add(sublist);
                        k++;
                        while (num[k] == num[k - 1] && k - 1 != j && k < l) {
                            k++;
                        }
                        l--;
                        while (num[l] == num[l + 1] && k < l) {
                            l--;
                        } 
                    }
                    else if (a + b + c + d < target) {
                        k++;
                        while (num[k] == num[k - 1] && k - 1 != j && k < l) {
                            k++;
                        }
                    }
                    else {
                        l--;
                        while (num[l] == num[l + 1] && k < l) {
                            l--;
                        } 
                    }
                }
                
                while (j + 1 < num.length - 2 && num[j + 1] == num[j]) {
                    j++;
                }
            }
            
            while (i + 1 < num.length - 3 && num[i + 1] == num[i]) {
                i++;
            }
        }
        
        return list;
    }
}