Leetcode 350. Intersection of Arrays II

Intersection of Arrays II

• Total accepted:23510
• Total submissions:56283
• Difficulty:easy

Given the arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1] , nums2 = [2, 2] , return [2, 2] .

Note:

• Each element of the result should appear as many times as it shows in both arrays.
• The result can be on any order.

Follow up:

• What if the given array is already sorted? How would optimize your algorithm?
• What if nums1' s size is small compared to nums2' s size? Which algorithm is better?
• What if elements of nums2 be stored on disk, and the memory is limited such so you cannot load all elements in To the memory at once?

Idea: and Leetcode 349. Intersection of the arrays similar, but note the difference of the return value.

In addition, I learned the usage of unordered_set and unordered_map .

Code:

Method One: Using Unordered_map, the complexity is lower than the method two.

1 classSolution {2  Public:3vector<int> Intersect (vector<int>& Nums1, vector<int>&nums2) {4unordered_map<int,int>m;5vector<int>Res;6          for(intI:NUMS1) m[i]++;7          for(inti:nums2) {8             if(m[i]-->0){9 Res.push_back (i);Ten             } One         } A         returnRes; -     } -};

Method Two:

1 classSolution {2  Public:3vector<int> Intersect (vector<int>& Nums1, vector<int>&nums2) {4 sort (Nums1.begin (), Nums1.end ());5 sort (Nums2.begin (), Nums2.end ());6vector<int>Res;7         intI=0, j=0;8          while(I<nums1.size () &&j<nums2.size ()) {9             //while (I+1<nums1.size () &&nums1[i]==nums1[i+1]) i++;//Remove duplicate elementsTen             //while (J+1<nums2.size () &&nums2[j]==nums2[j+1]) j + +; One             if(nums1[i]==Nums2[j]) { A Res.push_back (Nums2[j]); -i++; -J + +; the             } -             Else{ -nums1[i]>nums2[j]?j++:i++; -             } +         } -         returnRes; +     } A};

Leetcode 350. Intersection of Arrays II