Find K Pairs with Smallest Sums 解答

Question

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]

Solution

这道题的关键在于运用Heap这个数据结构。Heap push的时间复杂度是O(logN) 因此整体的时间复杂度是O(KlogK)

Python里如何自定义comparator? 可以传进去一个triple，第一个元素会被比较。

 1 from heapq import heappush, heappop
 2 class Solution:
 3     def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
 4         if not nums1 or not nums2:
 5             return []
 6         m, n = len(nums1), len(nums2)
 7         k = min(k, m * n)
 8         heap = []
 9         visited = set()
10         result = []
11         heappush(heap, (nums1[0] + nums2[0], 0, 0))
12         visited.add((0, 0))
13         while len(result) < k:
14             val = heappop(heap)
15             i, j = val[1], val[2]
16             result.append([nums1[i], nums2[j]])
17             if i + 1 < m and (i + 1, j) not in visited:
18                 heappush(heap, (nums1[i + 1] + nums2[j], i + 1, j))
19                 visited.add((i + 1, j))
20             if j + 1 < n and (i, j + 1) not in visited:
21                 heappush(heap, (nums1[i] + nums2[j + 1], i, j + 1))
22                 visited.add((i, j + 1))
23         return result