Clone an undirected graph. Each node in the graph contains a label
and a list of its neighbors
.
OJ's undirected graph serialization:
Nodes are labeled uniquely.
We use #
as a separator for each node, and ,
as a separator for node label and each neighbor of the node.
As an example, consider the serialized graph {0,1,2#1,2#2,2}
.
The graph has a total of three nodes, and therefore contains three parts as separated by #
.
- First node is labeled as
0
. Connect node0
to both nodes1
and2
. - Second node is labeled as
1
. Connect node1
to node2
. - Third node is labeled as
2
. Connect node2
to node2
(itself), thus forming a self-cycle.
Visually, the graph looks like the following:
1
/ \
/ \
0 --- 2
/ \
\_/
题目大意是给定一个无向图,可能有环,实现一个方法,deep clone这个图。
我的做法是采用BFS,从一个点开始,遍历它的邻居,然后加入队列,当队列非空,循环遍历,因为label是唯一的,采用map保存新生成的node,用label作为key。另外使用visited数组保存是否加入过队列,以免重复遍历。
Talk is cheap>>
public UndirectedGraphNode cloneGraph(UndirectedGraphNode root) {
HashSet<Integer> visited = new HashSet<>();
if (root==null)
return null;
List<UndirectedGraphNode> queue = new ArrayList<>();
HashMap<Integer,UndirectedGraphNode> map = new HashMap<>();
queue.add(root);
while (!queue.isEmpty()) {
UndirectedGraphNode node = queue.get(0);
if (map.get(node.label)==null) {
map.put(node.label, new UndirectedGraphNode(node.label));
}
UndirectedGraphNode tmp = map.get(node.label);
queue.remove(0); for (int i = 0; i < node.neighbors.size(); i++) {
int key = node.neighbors.get(i).label;
if (map.get(key)==null){
map.put(key,new UndirectedGraphNode(key));
}
tmp.neighbors.add(map.get(key));
visited.add(node.label);
if (!visited.contains(key)){
queue.add(node.neighbors.get(i));
visited.add(key);
}
}
}
return map.get(root.label);
}