【clone-graph】
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Clone an undirected graph. Each node in the graph contains alabeland a list of itsneighbors.
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 as0. Connect node0to both nodes1and2.
- Second node is labeled as1. Connect node1to node2.
- Third node is labeled as2. Connect node2to node2(itself), thus forming a self-cycle.
Visually, the graph looks like the following:
1 / \ / \ 0 --- 2 / \ \_/
题意:
克隆一个无向图,途中的每个节点包含一个标签及其邻居的列表
Oj对于无向图的序列化;
节点的标签是唯一的。
我们使用#作为节点的分割符,使用逗号,作为节点标签以及每个邻居之间的分割符;
例如,考虑序列化图0,1,2#1,2#2,2
上图包含三个几点,
第一个节点标签为0, 节点0与节点1、2相连;
第二个节点标签为1,节点1与节点2相连;
第三个节点为2,节点2与节点2相连,形成了一个环;
这道题主要是数据结构的应用吧,用一个队列存放途中还未被访问的节点,
用一个map存放克隆的新节点。对于队列中的节点,访问其邻居节点,如果该
邻居已经被克隆,那么只需要维护克隆节点之间的邻居关系即可,如果该邻居
尚未被克隆,那么需要new一个新节点作为它的克隆,并维护克隆后的邻居关系;
class Solution {public: map<int,UndirectedGraphNode*> record; UndirectedGraphNode *clone(UndirectedGraphNode *node) { UndirectedGraphNode *ngraph,*p; map<int,UndirectedGraphNode*>::iterator iter; p=node; iter=record.find(p->label); if(iter!=record.end()) return iter->second; ngraph=new UndirectedGraphNode(p->label); record.insert(map<int,UndirectedGraphNode*>::value_type(p->label,ngraph)); for(int i=0;i<p->neighbors.size();i++){ ngraph->neighbors.push_back(clone(p->neighbors[i])); } return ngraph; } UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node){ UndirectedGraphNode *ngraph; if(node==NULL) return NULL; record.clear(); ngraph=clone(node); return ngraph; }};
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