350. Intersection of Two Arrays II

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Given two arrays, write a function to compute their intersection.

Example 1:

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

Example 2:

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

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?

解法1：hashmap

解法2：排序

解法3：Built-in Intersection

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class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        vector<int>::iterator it =  set_intersection(begin(nums1), end(nums1), 
        begin(nums2), end(nums2), begin(nums1));
         nums1.erase(it, end(nums1));
        return nums1;
    }
};

set_intersection

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template<class InputIt1, class InputIt2, class OutputIt>
OutputIt set_intersection(InputIt1 first1, InputIt1 last1,
                          InputIt2 first2, InputIt2 last2,
                          OutputIt d_first)
{
    while (first1 != last1 && first2 != last2) {
        if (*first1 < *first2) {
            ++first1;
        } else  {
            if (!(*first2 < *first1)) {
                *d_first++ = *first1++;
            }
            ++first2;
        }
    }c
    return d_first;
}

双指针的实现。
set_intersection的功能：给定两个迭代器表示的元素范围，将这两个范围内的元素的交集拷贝到最后一个d_first参考给定的空间内。

参考文献

[1]set_intersection：https://en.cppreference.com/w/cpp/algorithm/set_intersection