My code:

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

public class Solution {
    private class entry {
        int[] key;
        int val;
        entry(int[] key, int val) {
            this.key = key;
            this.val = val;
        }
    }
    
    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        List<int[]> ret = new ArrayList<int[]>();
        if (nums1 == null || nums2 == null || k <= 0) {
            return ret;
        }
        
        PriorityQueue<entry> pq = new PriorityQueue<entry>(k, new Comparator<entry>() {
            public int compare(entry e1, entry e2) {
                return -1 * (e1.val - e2.val);             
            }
        });
        
        
        for (int i = 0; i < nums1.length; i++) {
            for (int j = 0; j < nums2.length; j++) {
                int[] key = new int[2];
                key[0] = nums1[i];
                key[1] = nums2[j];
                pq.offer(new entry(key, key[0] + key[1]));
                if (pq.size() > k) {
                    pq.poll();
                }
            }
        }
        
        while (!pq.isEmpty()) {
            ret.add(pq.poll().key);
        }
        
        Collections.sort(ret, new Comparator<int[]> () {
            public int compare(int[] o1, int[] o2) {
                return o1[0] + o1[1] - o2[0] - o2[1];
            }
        });
        return ret;
    }
}

其实和 top k frequent number 差不多的思路。
也是建一个priority queue存 map, value 作为优先级比较，key作为我们最后需要的输出。

Anyway, Good luck, Richardo! — 09/15/2016