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 = 3Return: [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 = 2Return: [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;}};