Leetcode题目：Intersection of Two Arrays II

题目：

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 num2'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?

题目解答：和上一题一样，这次不需要去重复。

代码如下：

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        sort(nums1.begin(), nums1.end());
        //vector<int>::iterator end_unique =  unique(nums1.begin(), nums1.end()); 
       // nums1.erase(end_unique, nums1.end());
        sort(nums2.begin(),nums2.end());
       // end_unique =  unique(nums2.begin(), nums2.end()); 
        //nums2.erase(end_unique, nums2.end());
        vector<int>::iterator nit1 = nums1.begin();
        vector<int>::iterator nit2 = nums2.begin();
        vector<int> res;
        while((nit1 != nums1.end())  && (nit2 != nums2.end()) )
        {
            if(*nit1 == *nit2)
            {
                res.push_back(*nit1);
                nit1++;
                nit2++;
            }
            else if(*nit1 < *nit2)
            {
                nit1++;
            }
            else
            {
                nit2++;
            }
            
        }
    return res;
    }
};

　　

考虑到Followup中提到的，对nums2进行排序不太好，使用哈希来编码实现，代码如下：

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int, int> num_map;
        vector<int> res;
        for(auto nit = nums1.begin(); nit != nums1.end(); nit++)
        {
            num_map[*nit] += 1;
        }
        for(auto nit = nums2.begin(); nit != nums2.end(); nit++)
        {
            if(num_map[*nit] >= 1)
            {
                num_map[*nit] -= 1;
                res.push_back(*nit);
            }
            
        }
        return res;
    }
};