350. Intersection of Two Arrays II

//HashMap Approach: Time: O(n), Space:O(n)  
  public int[] intersect(int[] nums1, int[] nums2) {
        if (nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0) {
            return new int[0];
        }
        
        List<Integer> result = new ArrayList<Integer>();
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        
        for (int i = 0; i < nums1.length; i++) {
            if (!map.containsKey(nums1[i])) {
                map.put(nums1[i], 1);
            } else {
                map.put(nums1[i], map.get(nums1[i]) + 1);
            }
        }
        
        for (int i = 0; i < nums2.length; i++) {
            if (map.containsKey(nums2[i])) {
                result.add(nums2[i]);
                map.put(nums2[i], map.get(nums2[i]) - 1);
                
                if (map.get(nums2[i]) == 0) {
                    map.remove(nums2[i]);
                }
            }
        }
        
        int[] ans = new int[result.size()];
        
        for (int i = 0; i < result.size(); i++) {
            ans[i] = result.get(i);
        }
        
        return ans;
    }
//Two pointer approach: Time: O(nlogn), Space: O(1)
//这种方法和349. Intersection of Two Arrays的区别是用一个用set hold结果，一个用list hold结果
    public int[] intersect(int[] nums1, int[] nums2) {
        if (nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0) {
            return new int[0];
        }
        
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        
        List<Integer> result = new ArrayList<Integer>();
        int i = 0; 
        int j = 0;
        
        while (i < nums1.length && j < nums2.length) {
            if (nums1[i] == nums2[j]) {
                result.add(nums1[i]);
                i++;
                j++;
            } else if (nums1[i] < nums2[j]) {
                i++;
            } else {
                j++;
            }
        }
        
        int[] ans = new int[result.size()];
        
        for (int k = 0; k < result.size(); k++) {
            ans[k] = result.get(k);
        }
        
        return ans;
    }