【LeetCode】350. Intersection of Two Arrays II

Difficulty:easy

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Description

https://leetcode.com/problems/intersection-of-two-arrays-ii/

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

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]

Note:

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

Follow up:

• What if the given array is already sorted? How would you optimize your algorithm?
• What if nums1's size is small compared to nums2's size? Which algorithm is better?
• 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?

Intuition

Method 1: HashMap

Method 2: Sorted arrays and two pointers

Solution

    //Method 1: HashMap
    public int[] intersect(int[] nums1, int[] nums2) {
        HashMap<Integer,Integer> map = new HashMap<>();
        ArrayList<Integer> list = new ArrayList<>();
        for(int i : nums1){
            if(map.containsKey(i)){
                map.put(i, map.get(i)+1);
            }else{
                map.put(i, 1);
            }
        }
        
        for(int i : nums2){
            if(map.containsKey(i) && map.get(i)>0){
                map.put(i, map.get(i)-1);
                list.add(i);
            }
        }
        int[] arr = new int[list.size()];
        for(int n=0; n<list.size(); n++){
            arr[n]=list.get(n);
        }
        return arr;
        
    }
    
    //Method 2: Sorted array & Two pointers
    public int[] intersect2(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        ArrayList<Integer> list = new ArrayList<>();
        int i=0, j=0;
        while(i<nums1.length && j<nums2.length){
            if(nums1[i]<nums2[j]){
                i++;
            }else if (nums1[i]>nums2[j]){
                j++;
            }else{
                list.add(nums1[i++]);
                j++;
            }
        }
        int[] arr = new int[list.size()];
        for(int n=0; n<list.size(); n++){
            arr[n]=list.get(n);
        }
        return arr;
    }

　　

Complexity

Method 1: 

Time complexity : O(N)

Space complexity : O(N)

Method 2:

Time complexity : O(NlogN)

Space complexity : O(1)

 

 More:【目录】LeetCode Java实现