共计 1952 个字符，预计需要花费 5 分钟才能阅读完成。
Given the head of a graph, return a deep copy (clone) of the graph. Each node in the graph contains a label (int) and a list (List[UndirectedGraphNode]) of its neighbors. There is an edge between the given node and each of the nodes in its neighbors.
OJ’s undirected graph serialization (so you can understand error output):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 node 0 to both nodes 1 and 2.Second node is labeled as 1. Connect node 1 to node 2.Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.
Visually, the graph looks like the following:
1
/ \
/ \
0 — 2
/ \
\_/
Note: The information about the tree serialization is only meant so that you can understand error output if you get a wrong answer. You don’t need to understand the serialization to solve the problem.
难度：medium
题目：给定一图的头结点，返回其深拷贝。每个结点包含一个标签及一组邻结点。在给定的结点与其邻结点之间都有一边。
思路：BFS
Runtime: 6 ms, faster than 19.61% of Java online submissions for Clone Graph.Memory Usage: 37.9 MB, less than 95.96% of Java online submissions for Clone Graph.
/**
* Definition for undirected graph.
* class UndirectedGraphNode {
* int label;
* List<UndirectedGraphNode> neighbors;
* UndirectedGraphNode(int x) {label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
* };
*/
public class Solution {
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
if (null == node) {
return null;
}
Queue<UndirectedGraphNode> queue = new LinkedList<>();
queue.add(node);
Map<UndirectedGraphNode, UndirectedGraphNode> nodes = new HashMap<>();
while (!queue.isEmpty()) {
UndirectedGraphNode tNode = queue.poll();
nodes.put(tNode, new UndirectedGraphNode(tNode.label));
for (UndirectedGraphNode n: tNode.neighbors) {
if (!nodes.containsKey(n)) {
queue.add(n);
}
}
}
for (Map.Entry<UndirectedGraphNode, UndirectedGraphNode> e: nodes.entrySet()) {
UndirectedGraphNode copyNode = e.getValue();
for (UndirectedGraphNode n : e.getKey().neighbors) {
copyNode.neighbors.add(nodes.get(n));
}
}
return nodes.get(node);
}
}
