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]
一刷
题解:构造heap, 首先将{nums1[0], nums2[I], 0}所有的I入队。0表示nums[2]里面的元素在队列中的len
如果nums1[i]被poll出来,读取常数,继续入队nums1[i], nums2[++len]
public class Solution {
public List kSmallestPairs(int[] nums1, int[] nums2, int k) {
PriorityQueue que = 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 0 && !que.isEmpty()){
int[] cur = que.poll();
res.add(new int[]{cur[0], cur[1]});
if(cur[2] == nums2.length-1) continue;
que.offer(new int[]{cur[0],nums2[cur[2]+1], cur[2]+1});
}
return res;
}
}