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:這道題是克隆一個一模一樣的圖。使用map來保存每個節點的克隆節點,使用隊列來遍歷圖中的每個節點——遍歷的結點出隊列,沒有遍歷的結點入隊列。
然后對於每個節點的鄰居結點,在map中查找,如果有,則將其push到該節點的克隆節點的鄰居結點(也是克隆);反之,則創建這個鄰居結點,然后作為該克隆節點的鄰居結點。
/** * 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]; } };