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]
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; } };