Restore the Array From Adjacent Pairs

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:


Solution:

construct a graph, then use bfs to restore the array

class Solution {
    public int[] restoreArray(int[][] adjacentPairs) {
        Map<Integer, List<Integer>> graph = new HashMap();
        for (int[] pair : adjacentPairs) {
            graph.computeIfAbsent(pair[0], k -> new ArrayList()).add(pair[1]);
            graph.computeIfAbsent(pair[1], k -> new ArrayList()).add(pair[0]);
        }
        int start = 1000000;
        for (int key : graph.keySet()) {
            if (graph.get(key).size() == 1) {
                start = key;
            }
        }
        Set<Integer> visited = new HashSet();
        Deque<Integer> queue = new ArrayDeque();
        queue.offer(start);
        visited.add(start);
        int i = 0;
        int[] res = new int[graph.size()];
        while (!queue.isEmpty()) {
            int curr = queue.poll();
            res[i ++] = curr;
            for (int next : graph.get(curr)) {
                if (visited.add(next)) {
                    queue.offer(next);
                }
            }
        }
        return res;
    }
}