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?
題目大意：
找出兩個數組中相同元素，相同元素的個數可以大於1.
思路：
將數組2的元素放入multiset中。
遍歷數組1，如果在multiset找到與數組1相等的元素，將該元素放入結果數組中，在multiset中刪除第一個和這個元素相當的元素。
代碼如下：
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;
}
};2016-08-13 14:03:35