Leetcode 350. Intersection of Two Arrays II

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?

解法一：

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> v;
        int len2 = nums2.size();
        for(int i = 0; i < len2; i++)
        {
            for(int j = 0; j < nums1.size(); j++)
            {
                 if(nums2[i] == nums1[j])
                 {
                    v.push_back(nums2[i]);
                    nums1.erase(nums1.begin()+j);
                    break;
                 }
            }
        }
        return v;
    }
};

解法二：

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

        if(nums1.empty() || nums2.empty())
        {
            return intersection;
        }

        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());


        int m = nums1.size();
        int n = nums2.size();

        int i = 0,j = 0;
        while(i < m && j < n)
        {
                 if(nums1[i] < nums2[j])
                 {
                     i++;
                 }
                 else if(nums1[i] > nums2[j])
                 {
                     j++;
                 }
                 else
                 {
                     intersection.push_back(nums1[i]);
                     i++;
                     j++;
                 }
         }


        return intersection;
    }
};

解法三：用hash表, 简单粗暴, 如果是有序的可以不用hash表, 同时扫描两个数组, 找相同的. 如果nums2在磁盘中, 用hash表无影响

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        if(nums1.size()==0 || nums2.size()==0) return vector<int>();
        vector<int> result;
        unordered_map<int, int> hash;
        for(auto val: nums1) hash[val]++;
        for(auto val: nums2)
        {
            if(hash.count(val) && hash[val]>0) result.push_back(val);
            hash[val]--;
        }
        return result;
    }
};

1. 如果不排序，O(mn)。 
2. 如果m和n都在合理范围内，先排序，再一个一个对比，时间复杂度O(nlgn + mlgm + m+n)。 
3. 如果m远小于n, 对n排序，m也排序（nlgn+mlgm+m+n），或者m不排序(nlgn + mn)。 这两种都差不多。也可以对m不排序，在n里做binary search，这样复杂度降低为nlgn+mlgn, 降很低。
4. 如果n很大，n只能存在disk上。只能把m load到内存，然后n一个一个的读进来，和m对比，这时m可以用hash存，这样复杂度就为O（n）了。