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

• Each element in the result should appear as many times as it shows in both arrays.
• The result can be in any order.

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        if(nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0)
            return new int[0];

        HashMap<Integer, Integer> map1 = new HashMap<>();
        HashMap<Integer, Integer> map2 = new HashMap<>();

        for(int n: nums1) {
            if(map1.containsKey(n))
                map1.put(n, 1 + map1.get(n));
            else
                map1.put(n, 1);
        }

        for(int n: nums2) {
            if(map2.containsKey(n))
                map2.put(n, 1 + map2.get(n));
            else
                map2.put(n, 1);
        }

        List<Integer> list = new ArrayList<Integer>();
        for(Integer key: map1.keySet()) {
            if(map2.containsKey(key)) {
                for(int i = 0; i < Math.min(map1.get(key), map2.get(key)); i++)
                    list.add(key);
            }
        }

        int[] res = new int[list.size()];
        for(int i = 0; i < res.length; i++)
            res[i] = list.get(i);

        return res;
    }
}

Follow up:

• What if the given array is already sorted? How would you optimize your algorithm?
  • Two pointers
• What if nums1's size is small compared to nums2's size? Which algorithm is better?
  • 交换nums1和nums2在代码中的位置
• What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?
  • 分批load