[leetcode]373. Find K Pairs with Smallest Sums找出和最小的前K对

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]

题意：

给定两个有序数组，允许从俩数组各选一个数组成数对。求和最小的前K对。

Solution： PriorityQueue

code

 1 class Solution {
 2     public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
 3         PriorityQueue<int[]> que = new PriorityQueue<>((a,b)->a[0]+a[1]-b[0]-b[1]);
 4         List<int[]> res = new ArrayList<>();
 5         if(nums1.length==0 || nums2.length==0 || k==0) return res;
 6         for(int i=0; i<nums1.length && i<k; i++) que.offer(new int[]{nums1[i], nums2[0], 0});
 7         while(k-- > 0 && !que.isEmpty()){
 8             int[] cur = que.poll();
 9             res.add(new int[]{cur[0], cur[1]});
10             if(cur[2] == nums2.length-1) continue;
11             que.offer(new int[]{cur[0],nums2[cur[2]+1], cur[2]+1});
12         }
13         return res;
14     }
15 }