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leetcode题目:Clone Graph
题目:
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:
1)遍历
2)用一张map存取旧指针和新指针,保持对应关系。
代码:
/** * 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; } UndirectedGraphNode* newnode = new UndirectedGraphNode(node->label); m_map.insert(make_pair(node,newnode)); checknewnode(node,newnode->neighbors); return newnode; } private: map<UndirectedGraphNode*,UndirectedGraphNode*>m_map; //判断这条顶点的邻居是否是新邻居 void checknewnode(UndirectedGraphNode* node,vector<UndirectedGraphNode*>&iivec) { vector<UndirectedGraphNode*>::iterator pos; for (pos=node->neighbors.begin();pos!=node->neighbors.end();pos++) { map<UndirectedGraphNode*,UndirectedGraphNode*>::iterator tmp = m_map.find(*pos); if (tmp==m_map.end()) { UndirectedGraphNode* newnode = new UndirectedGraphNode((*pos)->label); m_map.insert(make_pair(*pos,newnode)); checknewnode(*pos,newnode->neighbors); iivec.push_back(newnode); } else { iivec.push_back(tmp->second); } } } };
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