两个数组的交集 II

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

Example 1:

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

Example 2:

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?

题意：对于给定的两个数组，返回它们的交集，保留相同元素的出现相同次数。

思路：先用哈希表统计数组1各元素出现次数，然后遍历数组2，如果哈希表中该元素的出现次数>0，就把这个元素加入到结果数组，并把哈希表中该元素出现次数-1

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int,int> m;
        vector<int> ans;
        for(int i=0;i<nums1.size();i++){
            m[nums1[i]]++;
        }
        for(int i=0;i<nums2.size();i++){
            if(m[nums2[i]]){           //开始用m.count()判断不知道为啥不行
                ans.push_back(nums2[i]);
                m[nums2[i]]--;          
            }
        }
        return ans;
        
    }
};