349. Intersection of Two Arrays

Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must be unique and you may return the result in any order.

Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2]

Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [9,4]
Explanation: [4,9] is also accepted.

Solution:

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        Set<Integer> set = new HashSet<Integer>();

        for (int i = 0; i < nums1.length; i++){
            set.add(nums1[i]);
        }

        Set<Integer> resultSet = new HashSet<Integer>();
        for (int i = 0; i < nums2.length; i++){
            if (set.contains(nums2[i])){
                resultSet.add(nums2[i]);
            }
        }

        int[] result = new int[resultSet.size()];
        int i = 0;
        for (Integer item : resultSet){     // set 的用法要记住
            result[i] = item;
            i++;
        }
        return result;
    }
}
// TC: O(n)
// SC: O(n)

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        Set<Integer> set = new HashSet<Integer>();
        Set<Integer> set2 = new HashSet<Integer>();

        for (int i = 0; i < nums1.length; i++){
            set.add(nums1[i]);
        }

        for (int j = 0; j < nums2.length; j++){
            if (set.contains(nums2[j])){
                set2.add(nums2[j]);
            }
        }

        int[] result = new int[set2.size()];

        int k = 0;

        for (int ele : set2){
            result[k] = ele;
            k++;
        }

        return result;
    }
}

// TC: O(n)
// SC: O(n)