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
}