还原原数组

import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;
import java.util.stream.Collectors;

/**
 * Using Java Streams to process the array and find the original array `arr`.
 * The logic is the same but uses more functional programming style.
 */
public class Solution {
    public int[] recoverArray(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length / 2;
        for (int i = 1; i < nums.length; i++) {
            int k = (nums[i] - nums[0]) / 2;
            if (k > 0 && (nums[i] - nums[0]) % 2 == 0) {
                Set<Integer> lowerSet = new HashSet<>();
                Set<Integer> higherSet = Arrays.stream(nums).boxed().collect(Collectors.toSet());
                for (int num : nums) {
                    if (!lowerSet.contains(num - k)) {
                        lowerSet.add(num);
                    } else {
                        higherSet.remove(num);
                    }
                }
                if (lowerSet.size() == n && higherSet.size() == n) {
                    int[] result = lowerSet.stream().mapToInt(num -> num + k).toArray();
                    return result;
                }
            }
        }
        return new int[0];
    }
}