LeetCode(133) Clone Graph
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题目:复制一张无向图.
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 / \ \_/
解法:
遍历一遍原图即可,可以用bfs或dfs。容易错的是图有可能有多重边。另外邻接表中的元素是指针,所以如果访问某个节点n,它的邻居中有一个已经生成过的节点,要找到原来的这个节点,再把指向原来节点的指针放入n的邻接表中。否则,生成一个新节点,再把指向新节点的指针放入n的邻接表中。
代码:
class Solution {public: UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) { if(node==NULL) return NULL; map<int,UndirectedGraphNode*> hasCreated;//已生成的节点集合 UndirectedGraphNode *graph=new UndirectedGraphNode(node->label); hasCreated[node->label]=graph; dfs(graph,node,hasCreated); return graph; } void dfs(UndirectedGraphNode *newGraph,UndirectedGraphNode *oldGraph,map<int,UndirectedGraphNode*>& hasCreated) { for(int i=0;i<(oldGraph->neighbors).size();i++) { UndirectedGraphNode *neighbor=(oldGraph->neighbors)[i]; map<int,UndirectedGraphNode*>::iterator it=hasCreated.find(neighbor->label); if(it==hasCreated.end()) { UndirectedGraphNode *node=new UndirectedGraphNode(neighbor->label); (newGraph->neighbors).push_back(node); hasCreated[neighbor->label]=node; dfs(node,neighbor,hasCreated); } else { (newGraph->neighbors).push_back(it->second); } } }};
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