【Sort】350. Intersection of Two Arrays II

最新推荐文章于 2025-02-11 22:00:00 发布

原创最新推荐文章于 2025-02-11 22:00:00 发布 · 165 阅读

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本文介绍两种求两个数组交集的方法：使用哈希表和排序加双指针技术。哈希表解决方案的时间复杂度为O(m+n)，空间复杂度为O(m+n)；排序加双指针的时间复杂度为O(max(m,n)log(max(m,n)))，空间复杂度为O(m+n)。文章还探讨了不同场景下的最优解。

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?

解答：

被分类到了Sort的题目，看标签和Binary Search， Hash Table， Two Pointers都有关。

刚开始学Hash Table，解答来自Discuss:

m: nums1.size n: nums2.size

Hash table solution:
Time: O(m + n) Space: O(m + n)

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int, int> dict;
        vector<int> res;
        for(int i = 0; i < (int)nums1.size(); i++) dict[nums1[i]]++;
        for(int i = 0; i < (int)nums2.size(); i++)
            if(--dict[nums2[i]] >= 0) res.push_back(nums2[i]);
        return res;
    }
};

Hash table solution2:
Time: O(m + n) Space: O(m)

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int, int> dict;
        vector<int> res;
        for(int i = 0; i < (int)nums1.size(); i++) dict[nums1[i]]++;
        for(int i = 0; i < (int)nums2.size(); i++)
            if(dict.find(nums2[i]) != dict.end() && --dict[nums2[i]] >= 0) res.push_back(nums2[i]);
        return res;
    }
};

Sort and two pointers Solution:
Time: O(max(m, n) log(max(m, n))) Space: O(m + n)

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        int n1 = (int)nums1.size(), n2 = (int)nums2.size();
        int i1 = 0, i2 = 0;
        vector<int> res;
        while(i1 < n1 && i2 < n2){
            if(nums1[i1] == nums2[i2]) {
                res.push_back(nums1[i1]);
                i1++;
                i2++;
            }
            else if(nums1[i1] > nums2[i2]){
                i2++;
            }
            else{
                i1++;
            }
        }
        return res;
    }
};

还有一种解答用了set_intersection():

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        nums1.erase(set_intersection(nums1.begin(), nums1.end(), nums2.begin(), nums2.end(), nums1.begin()), nums1.end());//利用了set_intersection()，是std::set
        return nums1;
    }
};

这里补一下set_intersection这个函数，下面的代码描述了它的行为：

template <class InputIterator1, class InputIterator2, class OutputIterator>
  OutputIterator set_intersection (InputIterator1 first1, InputIterator1 last1,
                                   InputIterator2 first2, InputIterator2 last2,
                                   OutputIterator result)
{
  while (first1!=last1 && first2!=last2)
  {
    if (*first1<*first2) ++first1;
    else if (*first2<*first1) ++first2;
    else {
      *result = *first1;
      ++result; ++first1; ++first2;
    }
  }
  return result;
}

可见，最后第5个参数，也就是这里的nums1.begin()指向的是[2,2,0,0,0,0]的第一个0。