**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]
Constraints:
1 <= nums1.length, nums2.length <= 105
-109 <= nums1[i], nums2[i] <= 109
nums1 and nums2 both are sorted in ascending order.
1 <= k <= 1000
ac
note: java 8 lambda is much slower than anonymous class.
class Solution {
public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
List<int[]> res = new ArrayList<int[]>();
// edge cases
if (nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0 || k <= 0)
return res;
// object: [idx1, idx2]
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> {return nums1[a[0]] + nums2[a[1]] - nums1[b[0]] - nums2[b[1]];});
// init
for (int i = 0; i < nums1.length && i <= k; i++) {
pq.add(new int[]{i, 0});
}
// get result
while (!pq.isEmpty() && res.size() < k) {
int[] curr = pq.poll();
res.add(new int[]{nums1[curr[0]], nums2[curr[1]]});
if (curr[1] + 1 < nums2.length) {
curr[1]++;
pq.offer(curr);
}
}
return res;
}
}
/*
priority queue
add initial object to pq
poll one, add to res
move curr object's idx2 if idx2 < len2 -> add to pq
*/