Leetcode 350. Intersection of Two Arrays II

Posted 2020-07-12 琴影

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?

解法一：

 

 1 class Solution {
 2 public:
 3     vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
 4         vector<int> v;
 5         int len2 = nums2.size();
 6         for(int i = 0; i < len2; i++)
 7         {
 8             for(int j = 0; j < nums1.size(); j++)
 9             {
10                  if(nums2[i] == nums1[j])
11                  {
12                     v.push_back(nums2[i]);
13                     nums1.erase(nums1.begin()+j);
14                     break;
15                  }
16             }
17         }
18         return v;
19     }
20 };

View Code

 

解法二：

 1 class Solution {
 2 public:
 3     vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
 4         vector<int> intersection;
 5 
 6         if(nums1.empty() || nums2.empty())
 7         {
 8             return intersection;
 9         }
10 
11         sort(nums1.begin(), nums1.end());
12         sort(nums2.begin(), nums2.end());
13 
14 
15         int m = nums1.size();
16         int n = nums2.size();
17 
18         int i = 0,j = 0;
19         while(i < m && j < n)
20         {
21                  if(nums1[i] < nums2[j])
22                  {
23                      i++;
24                  }
25                  else if(nums1[i] > nums2[j])
26                  {
27                      j++;
28                  }
29                  else
30                  {
31                      intersection.push_back(nums1[i]);
32                      i++;
33                      j++;
34                  }
35          }
36 
37 
38         return intersection;
39     }
40 };

View Code

解法三：用hash表, 简单粗暴, 如果是有序的可以不用hash表, 同时扫描两个数组, 找相同的. 如果nums2在磁盘中, 用hash表无影响

 1 class Solution {
 2 public:
 3     vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
 4         if(nums1.size()==0 || nums2.size()==0) return vector<int>();
 5         vector<int> result;
 6         unordered_map<int, int> hash;
 7         for(auto val: nums1) hash[val]++;
 8         for(auto val: nums2)
 9         {
10             if(hash.count(val) && hash[val]>0) result.push_back(val);
11             hash[val]--;
12         }
13         return result;
14     }
15 };

View Code

1. 如果不排序，O(mn)。 
2. 如果m和n都在合理范围内，先排序，再一个一个对比，时间复杂度O(nlgn + mlgm + m+n)。 
3. 如果m远小于n, 对n排序，m也排序（nlgn+mlgm+m+n），或者m不排序(nlgn + mn)。 这两种都差不多。也可以对m不排序，在n里做binary search，这样复杂度降低为nlgn+mlgn, 降很低。
4. 如果n很大，n只能存在disk上。只能把m load到内存，然后n一个一个的读进来，和m对比，这时m可以用hash存，这样复杂度就为O（n）了。