nums1 and nums2 sorted in ascending order and an integer k.
(u,v)
(u1,v1),(u2,v2) ...(uk,vk)
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]
Credits:
Special thanks to @elmirap and @StefanPochmann for adding this problem and creating all test cases.
求数组中前k小的数的变种！
注意优先队列的这种声明方式
priority_queue<pair<int, int>, vector<pair<int, int>>,cmp> pq;
class Solution {
public:
struct cmp{
bool operator()(const pair<int,int>& a,const pair<int,int>&b){
return a.first + a.second < b.first + b.second;
}
};
vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
priority_queue<pair<int, int>, vector<pair<int, int>>,cmp> 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({ nums1[i], nums2[j] });
}
else{
if (nums1[i] + nums2[j] < pq.top().first + pq.top().second){
pq.pop();
pq.push({ nums1[i], nums2[j] });
}
}
}
}
vector<pair<int, int>> res;
while (!pq.empty()){
res.push_back(pq.top());
pq.pop();
}
return res;
}
};
