leetcode findkpairswithsmallestsums

import java.util.ArrayList;
import java.util.List;
import java.util.PriorityQueue;

public class  {
  
   * 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:
   *
   * Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
   * Output: [[1,2],[1,4],[1,6]]
   * Explanation: 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:
   *
   * Input: nums1 = [1,1,2], nums2 = [1,2,3], k = 2
   * Output: [1,1],[1,1]
   * Explanation: 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:
   *
   * Input: nums1 = [1,2], nums2 = [3], k = 3
   * Output: [1,3],[2,3]
   * Explanation: All possible pairs are returned from the sequence: [1,3],[2,3]
   */
  public List<int[]> solution(int[] nums1, int[] nums2, int k){
    PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0]+a[1]-b[0]-b[1]);
    List<int[]> ans = new ArrayList<>();
    for (int i=0; i<nums1.length && i<k; i++){
      pq.offer(new int[]{nums1[i], nums2[0], 0});
    }
    while(k > 0 && !pq.isEmpty()){
      int[] t = pq.poll();
      ans.add(t);
      if (t[2] == nums2.length)
        continue;
      pq.offer(new int[]{t[0], nums2[t[3]+1], t[3]+1});
      k--;
    }
    return ans;
  }



}