Skip to content

Latest commit

 

History

History
60 lines (44 loc) · 2.61 KB

File metadata and controls

60 lines (44 loc) · 2.61 KB

Medium


Solution

Quite similar to BFS/DFS.

  1. Construct a map of adjacent numbers.
  2. Determine starting number by finding the numbers which have only one adjacent number. Two candidates exist, while other one will be the end point.
  3. Traverse from the starting number. Implementing straightforward BFS/DFS will also work.

There is an integer array nums that consists of n unique elements, but you have forgotten it. However, you do remember every pair of adjacent elements in nums.

You are given a 2D integer array adjacentPairs of size n - 1 where each adjacentPairs[i] = [ui, vi] indicates that the elements ui and vi are adjacent in nums.

It is guaranteed that every adjacent pair of elements nums[i] and nums[i+1] will exist in adjacentPairs, either as [nums[i], nums[i+1]] or [nums[i+1], nums[i]]. The pairs can appear in any order.

Return the original array nums. If there are multiple solutions, return any of them.

 

Example 1:

Input: adjacentPairs = [[2,1],[3,4],[3,2]]
Output: [1,2,3,4]
Explanation: This array has all its adjacent pairs in adjacentPairs.
Notice that adjacentPairs[i] may not be in left-to-right order.

Example 2:

Input: adjacentPairs = [[4,-2],[1,4],[-3,1]]
Output: [-2,4,1,-3]
Explanation: There can be negative numbers.
Another solution is [-3,1,4,-2], which would also be accepted.

Example 3:

Input: adjacentPairs = [[100000,-100000]]
Output: [100000,-100000]

 

Constraints:

  • nums.length == n
  • adjacentPairs.length == n - 1
  • adjacentPairs[i].length == 2
  • 2 <= n <= 105
  • -105 <= nums[i], ui, vi <= 105
  • There exists some nums that has adjacentPairs as its pairs.