350. Intersection of Two Arrays II

题目：

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?

答案：

思路是先排序再比较

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