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 node0
to both nodes1
and2
. - Second node is labeled as
1
. Connect node1
to node2
. - Third node is labeled as
2
. Connect node2
to node2
(itself), thus forming a self-cycle.
Visually, the graph looks like the following:
1 / \ / \ 0 --- 2 / \ \_/
思路:这道题是克隆一个一模一样的图。使用map来保存每个节点的克隆节点,使用队列来遍历图中的每个节点——遍历的结点出队列,没有遍历的结点入队列。
然后对于每个节点的邻居结点,在map中查找,如果有,则将其push到该节点的克隆节点的邻居结点(也是克隆);反之,则创建这个邻居结点,然后作为该克隆节点的邻居结点。
/** * Definition for undirected graph. * struct UndirectedGraphNode { * int label; * vector<UndirectedGraphNode *> neighbors; * UndirectedGraphNode(int x) : label(x) {}; * }; */ class Solution { public: UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) { if(node==NULL) return NULL; map<UndirectedGraphNode *,UndirectedGraphNode *> clone; queue<UndirectedGraphNode *> q; clone[node]=new UndirectedGraphNode(node->label); q.push(node); while(!q.empty()) { UndirectedGraphNode *pCur=q.front(); q.pop(); for(int i=0;i<pCur->neighbors.size();i++) { UndirectedGraphNode *pNei=pCur->neighbors[i]; if(clone.find(pNei)!=clone.end()) { clone[pCur]->neighbors.push_back(clone[pNei]); } else { clone[pNei]=new UndirectedGraphNode(pNei->label); clone[pCur]->neighbors.push_back(clone[pNei]); q.push(pNei); } } } return clone[node]; } };