350.leetcode Intersection of Two Arrays II(easy)[数组 归并排序]

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?
题目的含义是要找两个数组中的完全交集，思想是将首先将两个数组排序，然后按照归并排序的思想用两个指针遍历两个数组得到相同的内容。

class Solution {
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
vector<int> result;
if(nums1.size()==0 || nums2.size() == 0) return result;
sort(nums1.begin(),nums1.end());
sort(nums2.begin(),nums2.end());
int i=0,j=0;
while(i != nums1.size() && j != nums2.size())
{
if(nums1[i] == nums2[j])
{
//cout<<nums1[i]<<endl;
result.push_back(nums1[i]);
i++;j++;
}
else if(nums1[i]<nums2[j])
{
i++;
}else
j++;
}
return result;
}
};