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问题描述
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
/ \_/
解决思路
因为节点的label值唯一，可以使用一个HashMap来记录<node.label, copy node>。
使用BFS来遍历所有的原始图上的节点，通过map查找到复制节点，然后添加neighbors。
程序
public class Solution {
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
if (node == null) {
return null;
}
HashMap<Integer, UndirectedGraphNode> map = new HashMap<>(); // label unique
Queue<UndirectedGraphNode> q = new LinkedList<>();
q.offer(node);
while (!q.isEmpty()) {
// bfs
int size = q.size();
for (int i = 0; i < size; i++) {
UndirectedGraphNode top = q.poll();
if (!map.containsKey(top.label)) {
map.put(top.label, new UndirectedGraphNode(top.label));
}
UndirectedGraphNode copy = map.get(top.label);
List<UndirectedGraphNode> neighbors = top.neighbors;
if (neighbors != null && neighbors.size() > 0) {
for (UndirectedGraphNode n : neighbors) {
if (!map.containsKey(n.label)) {
map.put(n.label, new UndirectedGraphNode(n.label));
q.offer(n);
}
copy.neighbors.add(map.get(n.label));
}
}
}
}
return map.get(node.label);
}
}
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原文地址：http://www.cnblogs.com/harrygogo/p/4728245.html
