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 = 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]

Solution：Heap

思路：
Time Complexity: O(klogk) Space Complexity: O(k)

Solution Code：

public class Solution {
    public List kSmallestPairs(int[] nums1, int[] nums2, int k) {
        PriorityQueue queue = new PriorityQueue<>((a, b) -> a[0] + a[1] - b[0] - b[1]);
        List res = new ArrayList<>();
        if(nums1.length == 0 || nums2.length == 0 || k == 0) {
            return res;
        }
        for(int i = 0; i < nums1.length && i < k; i++) {
            queue.offer(new int[]{nums1[i], nums2[0], 0});   // nums2[i], nums2[j], index for nums2
        }

        while(k-- > 0 && !queue.isEmpty()) {
            int[] cur = queue.poll();
            res.add(new int[]{cur[0], cur[1]});
            if(cur[2] == nums2.length - 1) continue;
            queue.offer(new int[]{cur[0], nums2[cur[2]+1], cur[2]+1});   // essential
        }
        return res;
    }
}