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更新时间：2022-10-03 13:48:33
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 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?
题目大意：
找出两个数组中相同元素，相同元素的个数可以大于1.
思路：
将数组2的元素放入multiset中。
遍历数组1，如果在multiset找到与数组1相等的元素，将该元素放入结果数组中，在multiset中删除第一个和这个元素相当的元素。
代码如下：
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class Solution {
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
vector<int> result;
multiset<int> set2;
for(int i = 0; i < nums2.size(); i++)
set2.insert(nums2[i]);
for(int i = 0; i < nums1.size(); i++)
{
if(set2.find(nums1[i]) != set2.end())
{
result.push_back(nums1[i]);
set2.erase(set2.find(nums1[i]));
}
}
return result;
}
}; |