350. Intersection of Two Arrays II

Leetcode Hash Table Two Pointers Binary Search Sort

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?

分析

两个数组的交集。这道题目是LeetCode 349. Intersection of Two Arrays的延伸，前者返回的交集中每个元素出现一次，这里要求中每个元素出现的次数应与元素在两个数组中出现的次数一致。那么一种非常直接的方法就是把前者的Set该成HashMap来记录每个元素出现的次数。

public int[] intersect(int[] nums1, int[] nums2) {
    Map<Integer, Integer> map = new HashMap<>();
    List<Integer> list = new ArrayList<>();
    for (int num : nums1) map.put(num, map.getOrDefault(num, 0) + 1);
    for (int num : nums2)
        if (map.containsKey(num) && map.get(num) > 0) {
            map.put(num, map.get(num) - 1);
            list.add(num);
        }
    int[] res = new int[list.size()];
    for (int i = 0; i < list.size(); i++)
        res[i] = list.get(i);
    return res;
}