【数据结构入门_数组】 Leetcode 350. 两个数组的交集 II

原题连接：Leetcode 350. Intersection of Two Arrays II

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]
1
2

Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
Explanation: [9,4] is also accepted.
1
2
3

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?

---

方法一：哈希表

思路：

用哈希表记录nums1中出现的元素的次数。然后遍历nums2中的元素，当元素在哈希表中时，添加到ans中，并且减少该元素出现的次数。

c++代码：

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        // 对较短的那个数组用哈希表来记录, 空间复杂度会降低
        if(nums1.size() > nums2.size())
            return intersect(nums2, nums1);
        
        // <值, 出现次数>
        unordered_map<int, int> mp;
        vector<int> ans;

        // 遍历nums1, 统计元素出现的次数到哈希表
        for(auto num1 : nums1){
            mp[num1]++;
        }

        // 遍历nums2, 找到在出现过的元素添加到ans, 并更新哈希表 
        for(auto num2 : nums2){
            if(mp.count(num2)){
                ans.push_back(num2);
                --mp[num2];
                if(mp[num2] == 0)
                    mp.erase(num2);
            }
        }

        return ans;
    }
};
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29

复杂度分析：

• 时间复杂度：O(m+n)，需要遍历两个数组的所有元素
• 空间复杂度：O(min(m+n))，哈希表的长度是两个数组中长度较小的那一个