Title
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?
题目大意:求两个集合的交集，这个和349不同的是，结果集要加入重复的，也就是说不能用set了，用map做
Solution
自己做的，ac
//cpp:
class Solution {
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
unordered_map<int,int> map1;
for (int i = 0; i < nums1.size(); i++) {
map1[nums1[i]]++;
}
vector<int> v;
for (int i = 0; i < nums2.size(); i++) {
if (map1.count(nums2[i])==1&&map1[nums2[i]]>0) {
v.push_back(nums2[i]);
map1[nums2[i]]--;
}
}
return v;
}
};
假如有序的话，就用two pointers做，参考solution
class Solution {
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
sort(nums1.begin(), nums1.end());
sort(nums2.begin(), nums2.end());
int n1 = (int)nums1.size(), n2 = (int)nums2.size();
int i1 = 0, i2 = 0;
vector<int> res;
while(i1 < n1 && i2 < n2){
if(nums1[i1] == nums2[i2]) {
res.push_back(nums1[i1]);
i1++;
i2++;
}
else if(nums1[i1] > nums2[i2]){
i2++;
}
else{
i1++;
}
}
return res;
}
};