Leetcode18题 【四数之和】递归解法

题目地址: ​​四数之和​​

递归思路代码如下:
但是执行耗费时间不合题意，这是缺陷

public static List<List<Integer>> fourSum(int[] nums, int target) {
//set用于对结果去重
        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) {
        //处理边界情况
        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);
        //第一种情况，nums[0]属于四元组中的一个元素
        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);
            }
        }
        //第二种情况,nums[0]不属于四元组中的一个元素
        List<List<Integer>> lists1 = fourSumToTarget(range, target, set, size);
        result.addAll(lists1);
        return result;
    }
}