# 2215. Find the Difference of Two Arrays

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`

### 解題

func findDifference(nums1 []int, nums2 []int) [][]int {
m1 := make(map[int]bool)
m2 := make(map[int]bool)
for _, n := range nums1 {
m1[n] = true
}
for _, n := range nums2 {
m2[n] = true
}
ans := [][]int{ []int{}, []int{} }
for n, _ := range m1 {
if !(m1[n] && m2[n]) {
ans[0] = append(ans[0], n)
}
}
for n, _ := range m2 {
if !(m1[n] && m2[n]) {
ans[1] = append(ans[1], n)
}
}
return ans
}