LeetCode 373 Find K Pairs with Smallest Sums

LeetCode

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.

Return the k pairs (u<sub>1</sub>, v<sub>1</sub>), (u<sub>2</sub>, v<sub>2</sub>), ..., (u<sub>k</sub>, v<sub>k</sub>) 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]

Constraints:

1 <= nums1.length, nums2.length <= 10<sup>5</sup> -10<sup>9</sup> <= nums1[i], nums2[i] <= 10<sup>9</sup>
nums1 and nums2 both are sorted in ascending order. 1 <= k <= 1000

注意事项：

1. 类似于BFS模板，只不过是将queue换成heap。
2. 将两数和加入到heap中，而不是下标的和(粗心)

Python代码：

OFFSET = [(0, 1), (1, 0)]
def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
	heap, res = [(nums1[0] + nums2[0], 0, 0)], []
	visited = set([0, 0])
	while heap:
		node = heapq.heappop(heap)
		res.append([nums1[node[1]], nums2[node[2]]])
		if len(res) == k:
			break
		for _dx, _dy in OFFSET:
			_x, _y = node[1] + _dx, node[2] + _dy
			if _x < len(nums1) and _y < len(nums2) and (_x, _y) not in visited:
				heapq.heappush(heap, (nums1[_x] + nums2[_y], _x, _y))
				visited.add((_x, _y))
	return res

算法分析：

时间复杂度为O(k)，空间复杂度O(1)。