Leetcode 18 problem [sum of four numbers] recursive solution

Title address : Sum of four numbers

The recursive thinking code is as follows :
But the time-consuming implementation is not to our satisfaction , This is a defect

public static List<List<Integer>> fourSum(int[] nums, int target) {
    
//set Used to de duplicate the result 
        Set<String> set = new HashSet<>();
        Arrays.sort(nums);
        System.out.println(Arrays.toString(nums));
        return fourSumToTarget(nums, target, set, 4);
    }

    private static List<List<Integer>> fourSumToTarget(int[] nums, int target, Set<String> set, int size) {
    
        // Dealing with border situations 
        if (nums.length < size || size == 0) {
    
            return Collections.emptyList();
        }
        if ((nums.length == 1 || size == 1) && nums[0] == target) {
    
            return Collections.singletonList(Collections.singletonList(nums[0]));
        }
        List<List<Integer>> result = new ArrayList<>();
        int[] range = Arrays.copyOfRange(nums, 1, nums.length);
        // Case one ,nums[0] Belongs to an element in a quadruple 
        List<List<Integer>> lists = fourSumToTarget(range, target - nums[0], set, size - 1);
        for (int i = 0; i < lists.size(); i++) {
    
            List<Integer> integers = lists.get(i);
            List<Integer> temp = new ArrayList<>(integers);
            temp.add(nums[0]);
            if (temp.size() == 4) {
    
                if (set.add(Arrays.toString(temp.toArray()))) {
    
                    result.add(temp);
                }
            } else {
    
                result.add(temp);
            }
        }
        // The second case ,nums[0] An element that does not belong to a quadruple 
        List<List<Integer>> lists1 = fourSumToTarget(range, target, set, size);
        result.addAll(lists1);
        return result;
    }
}