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:

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]

时间复杂度 o(klogk),o(n)

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class Solution {
    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        List<int[]> res = new ArrayList<>();
        
        if(nums1 == null || nums2== null || nums1.length == 0||nums2.length == 0 || k <0)
            return res;
        
        PriorityQueue<int[]> queue = new PriorityQueue<>((a,b) -> a[0]+a[1] - b[0]-b[1]);
        
        
        for(int i = 0; i < nums1.length && i < k;i++){
            queue.offer(new int[]{nums1[i],nums2[0],0});
        }
        
        while(k-- > 0 && !queue.isEmpty()){
            int[] cur = queue.poll();
            
            res.add(new int[]{cur[0],cur[1]});
            
            if(cur[2] == nums2.length-1)
                continue;
            
            queue.offer(new int[]{cur[0],nums2[cur[2]+1],cur[2]+1 });
        }
        
        return res;
    }
}