0350. Intersection of Two Arrays II

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

Description

Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.

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]
**Explanation:** [9,4] is also accepted.

Constraints:

• 1 <= nums1.length, nums2.length <= 1000

• 0 <= nums1[i], nums2[i] <= 1000

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?

ac1: map

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

        int[] big = nums1.length > nums2.length ? nums1 : nums2;
        int[] small = nums1.length > nums2.length ? nums2 : nums1;

        // record nums in smaller one
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < small.length; i++) {
            map.put(small[i], map.getOrDefault(small[i], 0) + 1);
        }

        List<Integer> res = new ArrayList<>();
        // compair with bigger one
        for (int i = 0; i < big.length; i++) {
            if (map.getOrDefault(big[i], 0) > 0) {
                res.add(big[i]);
                map.put(big[i], map.get(big[i]) - 1);
            }
        }

        // build result
        int[] result = new int[res.size()];
        for (int i = 0; i < result.length; i++) {
            result[i] = res.get(i);
        }

        return result;
    }
}

ac2: sort + 2 pointers

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

        Arrays.sort(nums1);
        Arrays.sort(nums2);


        List<Integer> res = new ArrayList<>();
        // 2 pointers
        int p1 = 0, p2 = 0;
        while (p1 < nums1.length && p2 < nums2.length) {
            if (nums1[p1] == nums2[p2]) {
                res.add(nums1[p1]);
                p1++;
                p2++;
            } else if (nums1[p1] > nums2[p2]) {
                p2++;
            } else if (nums1[p1] < nums2[p2]) {
                p1++;
            }
        }

        // build result
        int[] result = new int[res.size()];
        for (int i = 0; i < result.length; i++) {
            result[i] = res.get(i);
        }

        return result;
    }
}