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。

from heapq import heappush, heappop
class Solution:
    def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
        if not nums1 or not nums2:
            return []
        m, n = len(nums1), len(nums2)
        k = min(k, m * n)
        heap = []
        visited = set()
        result = []
        heappush(heap, (nums1[0] + nums2[0], 0, 0))
        visited.add((0, 0))
        while len(result) < k:
            val = heappop(heap)
            i, j = val[1], val[2]
            result.append([nums1[i], nums2[j]])
            if i + 1 < m and (i + 1, j) not in visited:
                heappush(heap, (nums1[i + 1] + nums2[j], i + 1, j))
                visited.add((i + 1, j))
            if j + 1 < n and (i, j + 1) not in visited:
                heappush(heap, (nums1[i] + nums2[j + 1], i, j + 1))
                visited.add((i, j + 1))
        return result