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]
      

這個題和之前的 Kth Smallest Element in a Sorted Matrix 很像，用最大堆來解決。

class Pair implements Comparable<Pair>{
	int x;
	int y;
	
	public Pair(int x, int y) {
		super();
		this.x = x;
		this.y = y;
	}

	@Override
	public int compareTo(Pair o) {
		// TODO Auto-generated method stub
		return -(this.x+this.y-o.x-o.y);
	}
}
public class Solution {
   public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
		List<int[]> result=new ArrayList<int[]>();
        PriorityQueue<Pair> queue=new PriorityQueue<Pair>();
        int m=nums1.length,n=nums2.length;
        Pair[] pairs=new Pair[m*n];
        if(k<1)
        	return result;
        for(int i=0;i<m;i++){
        	for(int j=0;j<n;j++){
            	pairs[i*n+j]=new Pair(nums1[i], nums2[j]);
            }
        }
        for(int i=0;i<m*n;i++){
        	if(i<k){
        		queue.offer(pairs[i]);
        	}else{
        		Pair peek = queue.peek();
        		if(pairs[i].compareTo(peek)>0){
        			queue.poll();
        			queue.offer(pairs[i]);
        		}
        	}
        }
        k=k>m*n?m*n:k;
        for(int i=0;i<k;i++){
        	Pair pair=queue.poll();
        	result.add(0, new int[]{pair.x,pair.y});
        }
        return result;
    }
}