思路：用两个map分别存储两个列表内容，然后再对照即可
class Solution {
public:
vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
unordered_map<int, int> mp1;
unordered_map<int, int> mp2;
int ans1 = 0, ans2 = 0;
for (int i : nums1) {
mp1[i]++;
}
for (int j : nums2) {
mp2[j]++;
}
for (int i : nums1) {
if (mp2[i]) {
ans1++;
}
}
for (int i : nums2) {
if (mp1[i]) {
ans2++;
}
}
vector<int> res;
res.push_back(ans1);
res.push_back(ans2);
return res;
}
};
优化：初始化用集合set并且直接放入即可
标签：int,res,ans2,ans1,++,vector,数组,公共,2956 From： https://www.cnblogs.com/mysteryily/p/18306307