Title Description

    Here are two arrays of integers nums1 and nums2 , Please return the intersection of two arrays as an array . Returns the number of occurrences of each element in the result , It should be consistent with the number of occurrences of elements in both arrays （ If the number of occurrences is inconsistent , Then consider taking the smaller value ）. You can ignore the order of the output results .
Example 1

 Input ：nums1 = [1,2,2,1], nums2 = [2,2]
 Output ：[2,2]

Example 2

 Input ：nums1 = [4,9,5], nums2 = [9,4,9,8,4]
 Output ：[4,9]

Tips

• 1 <= nums1.length, nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 1000

Their thinking

Please pay attention to the hint , Prompt Element values in 0~1000, Using the idea of hash table , Put the elements of a container (x) As key , The number of occurrences is taken as the value . Traverse the elements in the second container (y), Element of judgement y Whether it is in the hash table , If yes, add , And the value corresponding to its key -1.

// Array implementation 
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
    
	vector<int>ans;
	int a[1005]={
    0};// Be sure to initialize to 0, Readers think about why ？
	for(auto x:nums1)
		a[x]++;
	for(auto y:nums2)
		if(a[y]){
    
		ans.push_back(y);
		a[y]--;
	}
	return ans;	
}

Hash table implementation （ Array implementation will be more efficient ）

vector<int> intersect1(vector<int>& nums1, vector<int>& nums2) {
    
	vector<int>ans;
	unordered_map<int,int> aa;
	for(auto x:nums1)
		aa[x]++;
	
	for(auto y:nums2)
		if(aa[y]){
    
		ans.push_back(y);
		--aa[y];
	}
	return ans;	
}