Intersection of two Arrays II (intersection of two arrays)

Topic

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

the

The intersection of two arrays, which contains the maximum number of occurrences of each number in the intersection as much as possible. Solving

Ideas and 349 ideas the same http://blog.csdn.net/xunalove/article/details/79372497, but not to heavy, while Nums1[i] in Nums2, delete nums2 in a nums1[i] Elements. The Python code is implemented. However, in C + + vector, remove () is the deletion of all equal to Nums1[i in nums2, not desirable.
Therefore, another method is used to save the number of occurrences of each number in the NUMS1 by using the map, traversing nums2 and processing. C + + code

Class Solution {public
:
    vector<int> intersect (vector<int>& nums1, vector<int>& NUMS2) {

        vector<int>res;
        map<int,int>m;

        for (int i=0; i<nums1.size (); i++)
        {
            m[nums1[i]]++;
        }
        for (int i=0; i<nums2.size (); i++)
        {
            if (m[nums2[i]]>0)
            {
                res.push_back (nums2[i));
                m[nums2[i]]--
            }
        }
        return res;
    }
;

Python code

Class Solution (object):
    def intersect (self, NUMS1, NUMS2): ""
        : Type Nums1:list[int]
        : type NUMS2: List[int]
        : rtype:list[int]
        ""
        res = []
        for I in range (0, Len (nums1)):
            if Nums1[i] in NUMS2:
                Res.append (Nums1[i])
                nums2.remove (Nums1[i]) #删除nums2中该数
            If Len (nums2) ==0 or Len (nums1) ==0:
                return Res return

        Res

Python code is simpler