350. Intersection of two arrays II

Here are two arrays of integers nums1 and nums2 , Please return the intersection of two arrays as an array . Returns the number of occurrences of each element in the result , It should be consistent with the number of occurrences of elements in both arrays （ If the number of occurrences is inconsistent , Then consider taking the smaller value ）. You can ignore the order of the output results .

Example 1：

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

Example 2:

 Input ：nums1 = [4,9,5], nums2 = [9,4,9,8,4]
 Output ：[4,9]

Tips ：

1 <= nums1.length, nums2.length <= 1000
0 <= nums1[i], nums2[i] <= 1000

Advanced ：

What if the given array has been ordered ？ How will you optimize your algorithm ？
If nums1 Size ratio nums2 Small , Which method is better ？
If nums2 Elements of are stored on disk , Memory is limited , And you can't load all the elements into memory at once , What are you gonna do? ？

class Solution {
    
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
    
        sort(nums1.begin(),nums1.end());
        sort(nums2.begin(),nums2.end());
        int m = nums1.size();
        int n = nums2.size();
        int n1 = 0, n2 =0;
        vector<int> arr;
        while(n1 < m && n2 < n){
    
            if (nums1[n1] == nums2[n2]) {
    
                arr.push_back(nums1[n1]);
                n1++;
                n2++;
            } else if (nums1[n1] < nums2[n2]) {
    
                n1++;
            } else n2++;
        }
         return arr;
    }
};