题目链接:https://leetcode.com/problems/clone-graph/
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.
Example:
Input: {"$id":"1","neighbors":[{"$id":"2","neighbors":[{"$ref":"1"},{"$id":"3","neighbors":[{"$ref":"2"},{"$id":"4","neighbors":[{"$ref":"3"},{"$ref":"1"}],"val":4}],"val":3}],"val":2},{"$ref":"4"}],"val":1} Explanation: Node 1's value is 1, and it has two neighbors: Node 2 and 4. Node 2's value is 2, and it has two neighbors: Node 1 and 3. Node 3's value is 3, and it has two neighbors: Node 2 and 4. Node 4's value is 4, and it has two neighbors: Node 1 and 3.
Note:
- The number of nodes will be between 1 and 100.
- The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph.
- Since the graph is undirected, if node p has node q as neighbor, then node q must have node p as neighbor too.
- You must return the copy of the given node as a reference to the cloned graph.
思路一:
dfs:
AC 1ms 100% java:
/*
// Definition for a Node.
class Node {
public int val;
public List<Node> neighbors;
public Node() {}
public Node(int _val,List<Node> _neighbors) {
val = _val;
neighbors = _neighbors;
}
};
*/
class Solution {
Map<Integer,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.val))
return map.get(node.val);
Node newNode=new Node(node.val,new ArrayList<Node>());
map.put(node.val,newNode);
for(Node neighbor:node.neighbors){
newNode.neighbors.add(clone(neighbor));
}
return newNode;
}
}
方法二:BFS
/*
// Definition for a Node.
class Node {
public int val;
public List<Node> neighbors;
public Node() {}
public Node(int _val,List<Node> _neighbors) {
val = _val;
neighbors = _neighbors;
}
};
*/
class Solution {
public Node cloneGraph(Node node) {
if(node==null)
return null;
Queue<Node> queue=new LinkedList();
queue.add(node);
Map<Node,Node> map=new HashMap();
map.put(node,new Node(node.val,new ArrayList<Node>()));
while(!queue.isEmpty()){
Node temp=queue.poll();
for(Node neighbor:temp.neighbors){
if(!map.containsKey(neighbor)){
map.put(neighbor,new Node(neighbor.val,new ArrayList<Node>()));
queue.add(neighbor);
}
map.get(temp).neighbors.add(map.get(neighbor));
}
}
return map.get(node);
}
}