leetcode349. Intersection of Two Arrays

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].

Note:
Each element in the result must be unique.
The result can be in any order.

    public int[] intersection(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        List<Integer> nums3 = new ArrayList<Integer>();
            int i=0, j=0;
        while(i<nums1.length && j<nums2.length){
            if(nums1[i]==nums2[j])
            {
                if(!nums3.contains(nums1[i]))
                     nums3.add(nums1[i]);
                i++;
                j++;
             }
            else if(nums1[i]>nums2[j])
                j++;
            else
                i++;
        }
        int[] arr = new int[nums3.size()];
        for(int k=0;k<nums3.size();k++)
            arr[k]=nums3.get(k);
        return arr;
    }

    public int[] intersection2(int[] nums1, int[] nums2){
        if(nums1==null || nums2==null || nums1.length == 0 || nums2.length == 0){
            return new int[0];
        }
        int max = nums1[0], min = nums1[0];
        for(int n : nums1){
            if(n > max) max = n;
            else if(n < min) min = n;
        }
        
        boolean[] index = new boolean[max - min + 1];
        for(int n : nums1){
            index[n - min] = true;
        }
        
        int count = 0;
        int[] tmp = new int[Math.min(nums1.length, nums2.length)];
        for(int n : nums2){
            if(n>=min && n<=max && index[n-min]){
                tmp[count++] = n;
                index[n-min] =false;
            }
        }
        return count == tmp.length ? tmp : Arrays.copyOf(tmp, count);
    }