leetcode之Clone Graph

原题如下：

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 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
/ \
\_/

这道题看起来挺复杂的，仍是一个深度复制的问题，问题的关键是在复制一个节点时需要判断其邻居是否已被创建，在此可以采用深度优先搜索DFS，利用unordered_map 来保存已经创建的节点，其中key为节点的label，value为节点，在复制节点时首先判断节点是否为空，为空直接返回空节点（此条件处理一个节点都没有的情况），如果节点在map中，则直接返回map中的节点，否则创建新节点，并将新节点加入map中，接下来按照原节点的neighbor依次递归复制新节点的neighbor，最后返回当前节点。

UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
unordered_map<int,UndirectedGraphNode*>mp;
return clone(node,mp);
}
UndirectedGraphNode *clone(UndirectedGraphNode *node,unordered_map<int,UndirectedGraphNode *>&mp){
if(node == NULL)
return node;
if(mp.find(node->label) != mp.end())
return mp[node->label];
UndirectedGraphNode *nNode = new UndirectedGraphNode(node->label);
mp[nNode->label] = nNode;     //
for(int i = 0; i < node->neighbors.size(); i++){
nNode->neighbors.push_back(clone(node->neighbors[i],mp));
}
return nNode;

}

这里用到了unordered_map,它是一个无序的map，其底层实现基于hash算法，所以其查询速度很快，另外，在创建新节点后及时将节点存入了map中，然后才创建其neighbor，我现在有点儿疑惑的是假如只有一个节点，且其neighbor为其自身，这是怎样创建的呢？