题目
Clone an undirected graph. Each node in the graph contains a label and
a list of its neighbors.
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 #.
0.
Connect node 0 to both nodes 1 and 2.1.
Connect node 1 to node 2.2.
Connect node 2 to node 2 (itself),
thus forming a self-cycle.Visually, the graph looks like the following:
1
/ / 0 --- 2
/ \_/
这题与Copy with Random Pointer思路一致，都是通过纪录原节点和新节点的映射关系来构建新的数据结构，只要正确遍历完所有节点即可，如下代码中采用BFS进行遍历。
代码
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.Stack;
public class CloneGraph {
class UndirectedGraphNode {
int label;
ArrayList<UndirectedGraphNode> neighbors;
UndirectedGraphNode(int x) {
label = x;
neighbors = new ArrayList<UndirectedGraphNode>();
}
}
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
if (node == null) {
return null;
}
Stack<UndirectedGraphNode> stack = new Stack<UndirectedGraphNode>();
Map<Integer, UndirectedGraphNode> map = new HashMap<Integer, UndirectedGraphNode>();
UndirectedGraphNode root = new UndirectedGraphNode(node.label);
map.put(node.label, root);
stack.push(node);
while (!stack.isEmpty()) {
UndirectedGraphNode v = stack.pop();
UndirectedGraphNode newV = map.get(v.label);
ArrayList<UndirectedGraphNode> newNeighbors = new ArrayList<UndirectedGraphNode>();
for (UndirectedGraphNode w : v.neighbors) {
UndirectedGraphNode newW = null;
if (map.containsKey(w.label)) {
newW = map.get(w.label);
} else {
newW = new UndirectedGraphNode(w.label);
map.put(w.label, newW);
stack.push(w);
}
newNeighbors.add(newW);
}
newV.neighbors = newNeighbors;
}
return root;
}
}
原文：http://blog.csdn.net/perfect8886/article/details/19302193