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: 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：

因為要去重范舀，所以用TwoPointers instead of hashmap
思路：排序后，?for every element, and then for every the other element, use two pointers to find other two, with skipping process
Time Complexity: O(N) Space Complexity: O(N)
Note：如果是返回位置index的話贬丛，排序＋two pointers 位置改變了 還需要原序列上找 index步势，不如用hashmap

Solution Code：

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if(nums == null || nums.length == 0) return result;
        Arrays.sort(nums);
        
        if (4 * nums[0] > target || 4 * nums[nums.length - 1] < target)  // early not found
            return result;
        
        for(int i = 0; i < nums.length - 3; i++) {
            if(i != 0 && nums[i] == nums[i - 1]) continue;  //skip num1 dup 
            for(int j = i + 1; j < nums.length - 2; j++) {
                if((j != i + 1) && nums[j] == nums[j - 1]) continue;   //skip num2 dup 
                int target2sum = target - nums[i] - nums[j];
                
                // two pointers
                int start = j + 1, end = nums.length - 1;
                while(start < end) {
                    if(nums[start] + nums[end] == target2sum) { 
                        result.add(Arrays.asList(nums[i], nums[j], nums[start], nums[end]));
                        while(start < end && nums[start] == nums[start + 1]) start++;  // skip num3 dup 
                        while(start < end && nums[end] == nums[end - 1]) end--;    // skip num4 dup 
                        start++; 
                        end--;
                    }
                    else if(nums[start] + nums[end] < target2sum) {
                        start++;
                    }
                    else {
                        end--;
                    }
                }
                
            }
        }
        return result;
    }
}