[Leetcode] 350. Intersection of Two Arrays II

Description:
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?

Language: C++

解法一：
这种找到两个数组中重复元素的次数是典型是键值对应关系，用map来实现。注意map声明的时候要声明两个参数。

set和map中不允许重复元素但是vector可以。由于要存入重复的元素，因此直接用vector存就可。
Time: $O(nlogn)$
Space: $O(n)$

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        map<int,int> record;

        for(int i = 0; i < nums1.size(); i++)  
            record[nums1[i]]++;

        vector<int> res;
        for(int i = 0; i < nums2.size(); i++){

            if(record[nums2[i]] > 0){
                res.push_back(nums2[i]);
                record[nums2[i]]--;
            }
        }
        return res;
    }
};