373. Find K Pairs with Smallest

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

heap

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