Leetcode- intersection of two arrays ii- simple

title ：350 Intersection of two arrays II- Simple

subject

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? ？

Code Java

public int[] intersection(int[] nums1, int[] nums2) {
    
    HashMap map1 = new HashMap();
    HashMap map2 = new HashMap();
    HashMap map3 = new HashMap();
    //  hold nums1  Fill to  map1 k yes num v  Is the number of occurrences 
    for (int i = 0; i < nums1.length; i++) {
    
        Object o1 = map1.get(nums1[i]);
        if (o1 != null) {
    
            int count = (int) o1 + 1;
            map1.put(nums1[i], count);
        } else {
    
            map1.put(nums1[i], 1);
        }
    }
    //  hold nums2  Fill to  map2 k yes num v  Is the number of occurrences 
    for (int i = 0; i < nums2.length; i++) {
    
        Object o2 = map2.get(nums2[i]);
        if (o2 != null) {
    
            int count = (int) o2 + 1;
            map2.put(nums2[i], count);
        } else {
    
            map2.put(nums2[i], 1);
        }
    }
    //  Traverse nums2 Of map2  mapping  nums1 Of map1  Have the same elements   Put in map3  take v Put small elements into map1
    Set set = map2.entrySet();
    for (Object o : set) {
    
        Map.Entry entry = (Map.Entry) o;
        Object key = entry.getKey();
        int k = (int) key;
        Object o1 = map1.get(k);
        if (o1 != null) {
    
            int x1 = (int) o1;
            int x2 = (int) entry.getValue();
            if (x1 > x2){
    
                map3.put(k, x2);
            } else {
    
                map3.put(k, x1);
            }
        }
    }
    //  hold map1 Put the elements in   Array 
    ArrayList<Integer> arrayList = new ArrayList<Integer>();
    set = map3.entrySet();
    for (Object o : set) {
    
        Map.Entry entry = (Map.Entry) o;
        int sum = (int) entry.getValue();
        for (int i = 0; i < sum; i++) {
    
            arrayList.add((int)entry.getKey());
        }
    }
    int[] result = new int[arrayList.size()];
    for (int i = 0; i < result.length; i++) {
    
        result[i] = arrayList.get(i);
    }
    return result;
}

//  Advanced 
		public int[] intersection1(int[] nums1, int[] nums2) {
    
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        //  Double pointer 
        ArrayList<Integer> arrayList = new ArrayList<Integer>();
        int n = nums1.length;
        int m = nums2.length;
        for (int i = 0, j = 0; i < n && j < m;) {
    
            if (nums1[i] == nums2[j]) {
    
                arrayList.add(nums1[i]);
                i ++;
                j ++;
            } else if (nums1[i] > nums2[j]) {
    
                j ++;
            } else {
    
                i++;
            }
        }
        int[] result = new int[arrayList.size()];
        for (int i = 0; i < result.length; i++) {
    
            result[i] = arrayList.get(i);
        }
        return result;
    }