# 8ms C++ solution using sorting

• ``````class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {

if (nums1.empty() || nums2.empty()) {
return vector<int>();
}

sort(nums1.begin(), nums1.end());
sort(nums2.begin(), nums2.end());

return nums2[0] < nums1[0] ? intersection2(nums1, nums2) : intersection2(nums2, nums1);
}
vector<int> intersection2(vector<int>& nums1, vector<int>& nums2) {

vector<int> res;

size_t pos1 = 0;
size_t pos2 = 0;

size_t size1 = nums1.size();
size_t size2 = nums2.size();
while (true) {
while (pos2 < size2 && nums2[pos2] < nums1[pos1]) {
++pos2;
}
if (pos2 == size2) {
break;
}
if (nums2[pos2] == nums1[pos1]) {
res.push_back(nums2[pos2++]);
while (pos2 < size2 && nums2[pos2]==nums2[pos2-1]) {
// Move forward if we have duplicates
++pos2;
}
}
++pos1;
while (pos1 < size1 && nums1[pos1]==nums1[pos1-1]) {
// Move forward if we have duplicates
++pos1;
}

if (pos1 == size1) {
break;
}
}
return res;
}
};
``````

For Intersection of Two Arrays II, we just need to remove the two while loops where we move forward in the case of duplicates

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