133. Clone Graph - Medium

Given a reference of a node in a connected undirected graph.

Return a deep copy (clone) of the graph.

Each node in the graph contains a val (int) and a list (List[Node]) of its neighbors.

class Node {
    public int val;
    public List<Node> neighbors;
}

 

Test case format:

For simplicity sake, each node's value is the same as the node's index (1-indexed). For example, the first node with val = 1, the second node with val = 2, and so on. The graph is represented in the test case using an adjacency list.

Adjacency list is a collection of unordered lists used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.

The given node will always be the first node with val = 1. You must return the copy of the given node as a reference to the cloned graph.

 

Example 1:

Input: adjList = [[2,4],[1,3],[2,4],[1,3]]
Output: [[2,4],[1,3],[2,4],[1,3]]
Explanation: There are 4 nodes in the graph.
1st node (val = 1)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
2nd node (val = 2)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).
3rd node (val = 3)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
4th node (val = 4)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).

Example 2:

Input: adjList = [[]]
Output: [[]]
Explanation: Note that the input contains one empty list. The graph consists of only one node with val = 1 and it does not have any neighbors.

Example 3:

Input: adjList = []
Output: []
Explanation: This an empty graph, it does not have any nodes.

Example 4:

Input: adjList = [[2],[1]]
Output: [[2],[1]]

 

Constraints:

  • 1 <= Node.val <= 100
  • Node.val is unique for each node.
  • Number of Nodes will not exceed 100.
  • There is no repeated edges and no self-loops in the graph.
  • The Graph is connected and all nodes can be visited starting from the given node.

 

M1: BFS, time: O(V + E), space: O(V)

class Solution {
    public Node cloneGraph(Node node) {
        if(node == null) {
            return node;
        }
        Map<Node, Node> map = new HashMap<>();
        map.put(node, new Node(node.val, new ArrayList<>()));
        Queue<Node> q = new LinkedList<>();
        q.offer(node);
        
        while(!q.isEmpty()) {
            Node cur = q.poll();
            for(Node nei : cur.neighbors) {
                if(!map.containsKey(nei)) {
                    map.put(nei, new Node(nei.val, new ArrayList<>()));
                    q.offer(nei);
                }
                map.get(cur).neighbors.add(map.get(nei));
            }
        }
        return map.get(node);
    }
}

M2: DFS, time: O(V + E), space: O(V)

class Solution {
    Map<Node, Node> map = new HashMap<>();
    
    public Node cloneGraph(Node node) {
        return clone(node);
    }
    
    public Node clone(Node node) {
        if(node == null) {
            return null;
        }
        if(map.containsKey(node)) {
            return map.get(node);
        }
        Node newNode = new Node(node.val, new ArrayList<>());
        map.put(node, newNode);
        for(Node nei : node.neighbors) {
            newNode.neighbors.add(clone(nei));
        }
        return newNode;
    }
}

 

posted @ 2020-10-29 13:45  fatttcat  阅读(78)  评论(0编辑  收藏  举报