Leetcode 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.

nums1和nums2都是有序的，那么结果需要的k个组合就是按照俩俩的sum的大小来排序的

基本思路是 先把nums1里面所有元素和nums2[0]的组合全部放进priorityqueue里面

再每次取出最小的一对的同时， 加入nums2[]里的其他元素

public class Solution {    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {        PriorityQueue<int[]> queue = new PriorityQueue<>((a, b) -> a[0] + a[1] - b[0] - b[1]);        List<int[]> result = new ArrayList<>();                        if(nums1 == null || nums2 == null || nums1.length == 0 || nums2.length == 0 || k == 0){            return result;        }         int i = 0;        while(i < nums1.length && i < k){            queue.offer(new int[]{nums1[i++], nums2[0], 0});//初始化的时候第三个element都是0 代表nums2的第0个         }                                 while(k-- > 0 && !queue.isEmpty()){            int[] present = queue.poll();            int[] tmp = new int[]{present[0], present[1]};            result.add(tmp);            if(present[2] == nums2.length - 1) continue;            queue.offer(new int[]{present[0], nums2[present[2] + 1], present[2] + 1});//第三个element用来计数 依次加入nums2里面的元素        }        return result;    }}