[LeetCode] Find the Difference of Two Arrays

題目：Find the Difference of Two Arrays
難度：Easy

Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:

• answer[0]is a list of all distinct integers in nums1 which are not present in nums2.
• answer[1] is a list of all distinct integers in nums2 which are not present in nums1.

Note that the integers in the lists may be returned in any order.

Example 1:

Input: nums1 = [1,2,3], nums2 = [2,4,6]
Output: [[1,3],[4,6]]
Explanation:
For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3].
For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums2. Therefore, answer[1] = [4,6].

Example 2:

Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]
Output: [[3],[]]
Explanation:
For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3].
Every integer in nums2 is present in nums1. Therefore, answer[1] = [].

Constraints:

• 1 <= nums1.length, nums2.length <= 1000
• -1000 <= nums1[i], nums2[i] <= 1000

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題目有說回傳的 list 可以為任何排序，故可以使用 HashSet 來解題

public class Solution {
    public IList<IList<int>> FindDifference(int[] nums1, int[] nums2) {
        ISet<int> SetNums1 = new HashSet<int>(nums1);
        ISet<int> SetNums2 = new HashSet<int>(nums2);

        SetNums1.ExceptWith(nums2);
        SetNums2.ExceptWith(nums1);

        IList<int> list1 = SetNums1.ToList();
        IList<int> list2 = SetNums2.ToList();

        return new List<IList<int>>{list1, list2};
    }
}