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]
from heapq import heappop, heappush
class Solution:
def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
res, q = [], []
if len(nums1) == 0 or len(nums2) == 0 or k == 0:
return res
for i in range(k):
if i >= len(nums1):
break
heappush(q, (nums1[i] + nums2[0], nums1[i], 0))
while k and q:
cur = heappop(q)
res.append([cur[1], cur[0] - cur[1]])
k -= 1
if cur[2] == len(nums2) - 1:
continue
heappush(q, (cur[1] + nums2[cur[2] + 1], cur[1], cur[2] + 1))
return res
class Solution {
public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
PriorityQueue<int[]> pq = 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++) {
// 可能最小值的candidates, 和当前指向nums2的位置
pq.offer(new int[]{nums1[i], nums2[0], 0});
}
for (int i = 0; i < k && !pq.isEmpty(); i++) {
int[] min = pq.poll();
res.add(new int[]{min[0], min[1]});
if (min[2] == nums2.length - 1) {
continue;
}
// 每次弹出一个就把它指向的下一个当作候选放入pq
pq.offer(new int[]{min[0], nums2[min[2] + 1], min[2] + 1});
}
return res;
}
}