Leetcode Everyday: 350. Intersection of Two Arrays II

jimhust

于 2016-05-21 20:01:50 发布

阅读量197

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/jimhust/article/details/51472044

leetcode 专栏收录该内容

69 篇文章

订阅专栏

本文介绍了一种计算两个整数数组交集的算法实现。通过使用哈希映射记录第一个数组中元素出现的频率，然后遍历第二个数组来确定交集元素及其重复次数。文章包含完整的Java代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

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 num2'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?

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        HashMap<Integer, Integer> hm = new HashMap<Integer,Integer>();
        for(int i: nums1){
            hm.put(i,hm.getOrDefault(i,0)+1);
        }
        ArrayList<Integer> arrl = new ArrayList<Integer>();
        
        for(int i: nums2){
            if(hm.containsKey(i)){
                arrl.add(i);
                int n = hm.get(i)-1;
                if(n==0) hm.remove(i);
                else hm.put(i,n);
            }
        }
        int ret[] = new int[arrl.size()];
        for(int i=0;i<arrl.size();i++){
            ret[i] = arrl.get(i);
        }
        return ret;
        
    }
}