leetcode -- Clone Graph

不要晃荡，找准方向

[问题描述]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
/ \
\_/

[解题思路]
深度优先遍历

UndirectedGraphNode *Solution::cloneGraph(UndirectedGraphNode *node){
if (node == NULL)
return NULL;
map<UndirectedGraphNode*, UndirectedGraphNode*> visited;
queue<UndirectedGraphNode*> bfs_q;
visited[node] = new UndirectedGraphNode(node->label);
bfs_q.push(node);
while(!bfs_q.empty()){
UndirectedGraphNode* tmp = bfs_q.front(); bfs_q.pop();
for (auto k : tmp->neighbors){
if (visited.find(k) == visited.end()){
UndirectedGraphNode* t = new UndirectedGraphNode(k->label);
visited[k] = t;
visited[tmp]->neighbors.push_back(t);
bfs_q.push(k);
}
else{
visited[tmp]->neighbors.push_back(visited[k]);
}
}
}
return visited[node];
}