350. Intersection of Two Arrays II
- Total Accepted: 23510
- Total Submissions: 56283
- Difficulty: Easy
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 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?
思路：和Leetcode 349. Intersection of Two Arrays类似，但注意返回值的区别。
另外学习了unordered_set和unordered_map的用法。
代码：
方法一：利用unordered_map，复杂度比方法二低。
1 class Solution { 2 public: 3 vector<int> intersect(vector<int>& nums1, vector<int>& nums2) { 4 unordered_map<int,int> m; 5 vector<int> res; 6 for(int i:nums1) m[i]++; 7 for(int i:nums2){ 8 if(m[i]-->0){ 9 res.push_back(i); 10 } 11 } 12 return res; 13 } 14 };
方法二：
1 class Solution { 2 public: 3 vector<int> intersect(vector<int>& nums1, vector<int>& nums2) { 4 sort(nums1.begin(),nums1.end()); 5 sort(nums2.begin(),nums2.end()); 6 vector<int> res; 7 int i=0,j=0; 8 while(i<nums1.size()&&j<nums2.size()){ 9 //while(i+1<nums1.size()&&nums1[i]==nums1[i+1]) i++;//去除重复元素 10 //while(j+1<nums2.size()&&nums2[j]==nums2[j+1]) j++; 11 if(nums1[i]==nums2[j]){ 12 res.push_back(nums2[j]); 13 i++; 14 j++; 15 } 16 else{ 17 nums1[i]>nums2[j]?j++:i++; 18 } 19 } 20 return res; 21 } 22 };