leetcode.373. Find K Pairs with Smallest Sums

You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k.

Define a pair (u,v) which consists of one element from the first array and one element from the second array.

Find the k pairs (u1,v1),(u2,v2)
...(uk,vk) with the smallest
sums.

Example 1:

Given nums1 = [1,7,11], nums2 = [2,4,6],  k = 3

Return: [1,2],[1,4],[1,6]

The first 3 pairs are returned from the sequence:
[1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]

Example 2:

Given nums1 = [1,1,2], nums2 = [1,2,3],  k = 2

Return: [1,1],[1,1]

The first 2 pairs are returned from the sequence:
[1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]

Example 3:

Given nums1 = [1,2], nums2 = [3],  k = 3

Return: [1,3],[2,3]

All possible pairs are returned from the sequence:
[1,3],[2,3]

class Solution {
private:
struct mycompare{
bool operator()(pair<int, int>& p1, pair<int, int>& p2){
return p1.first + p1.second < p2.first + p2.second;
}
};
public:
vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
vector<pair<int, int>> res;
priority_queue<pair<int,int>, vector<pair<int, int> >, mycompare> pq;
for(int i = 0; i < min((int)nums1.size(), k); i++){
for(int j = 0; j < min((int)nums2.size(), k); j++){
if(pq.size() < k)
pq.push(make_pair(nums1[i], nums2[j]));
else if(nums1[i] + nums2[j] < pq.top().first + pq.top().second){
pq.push(make_pair(nums1[i], nums2[j]));
pq.pop();
}
}
}
while(!pq.empty()){
res.push_back(pq.top());
pq.pop();
}
return res;
}
};