動機

要會看input決定要用什麼做法

Problem

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 (u1, v1), (u2, v2), ..., (uk, vk) with the smallest sums.

 

Example 1:

Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3Output: [[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 = 2Output: [[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 = 3Output: [[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

Sol

先生完所有組合，再挑

class Solution:
    def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
        q = []
        for n1 in nums1:
            for n2 in nums2:
                heappush(q,(n1+n2, (n1,n2)))
        
        ret = []
        for _ in range(k):
            if q:
                tmp = heappop(q)
                ret.append(tmp[1])
            else:
                break
        return ret

Sol: 四連棋

用一個array紀錄nums1的每個index的要從哪個nums2的index開始找

這樣就可以一次一次找出最小的pair

class Solution:
    def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
        idx_for_num1_to_start = [0]*len(nums1)
        ret = []
        for _ in range(min(k,len(nums1)*len(nums2))):
            cnt = math.inf
            minIdx = 0
            for i in range(len(nums1)):
                now = idx_for_num1_to_start[i]
                if now < len(nums2) and cnt >= nums1[i]+nums2[now]:
                    minIdx = i
                    cnt = nums1[i]+nums2[now]
            ret.append([nums1[minIdx],nums2[idx_for_num1_to_start[minIdx]]])
            idx_for_num1_to_start[minIdx] += 1
        return ret