QUESTION:
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
EXPLANATION:
easy的题目, 两个for循环可以搞定.
SOLUTION:
class Solution {
func findDifference(_ nums1: [Int], _ nums2: [Int]) -> [[Int]] {
var result:[[Int]] = []
result.append([])
result.append([])
for int in nums1 {
if !nums2.contains(int) && !result[0].contains(int) {
result[0].append(int)
}
}
for int in nums2 {
if !nums1.contains(int) && !result[1].contains(int) {
result[1].append(int)
}
}
return result
}
}