Given two arrays, write a function to compute their intersection. **Example:**Given nums1 = [1, 2, 2, 1] , nums2 = [2, 2] , return [2, 2] 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? Subscribe to see which companies asked this question.

The title

Compute the intersection of two arrays

The sample nums1 = [1, 2, 2, 1], nums2 = [2, 2], return to [2, 2].

Analysis of the

This is an extension of the previous problem, but this time we’re going to keep track of the number of elements that are repeated, so we’re going to use a hash table where the keys are repeated, and the values are repeated. Use the hashset method

code

public class Solution {
    /**
     * @param nums1 an integer array
     * @param nums2 an integer array
     * @return an integer array
     */
    public int[] intersection(int[] nums1, int[] nums2) {
        // Write your code here
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int i = 0; i < nums1.length; ++i) {
            if (map.containsKey(nums1[i]))
                map.put(nums1[i], map.get(nums1[i]) + 1); 
            else
                map.put(nums1[i], 1);
        }

        List<Integer> results = new ArrayList<Integer>();

        for (int i = 0; i < nums2.length; ++i)
            if (map.containsKey(nums2[i]) &&
                map.get(nums2[i]) > 0) {
                results.add(nums2[i]);
                map.put(nums2[i], map.get(nums2[i]) - 1); 
            }

        int result[] = new int[results.size()];
        for(int i = 0; i < results.size(); ++i)
            result[i] = results.get(i);

        returnresult; }}Copy the code