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]
留坑，后面补充其他解法
public class Solution { public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) { List<int[]> res = new ArrayList<int[]>(); PRiorityQueue<int[]> queue = new PriorityQueue<>(k, new Comparator<int[]>(){ @Override public int compare(int[] o1, int[] o2){ return o1[0] + o1[1] - o2[0] - o2[1]; } }); for (int i = 0; i < nums1.length; i++) { for (int j = 0; j < nums2.length; j++) { queue.add(new int[]{nums1[i], nums2[j]}); } } while (k-- > 0) { int[] tmp = queue.poll(); if(tmp == null) break; res.add(tmp); } return res; }}新闻热点
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