LeetCode-133-Clone Graph
Posted 无名路人甲
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算法描述:
Given the head of a graph, return a deep copy (clone) of the graph. Each node in the graph contains a label
(int
) and a list (List[UndirectedGraphNode]
) of its neighbors
. There is an edge between the given node and each of the nodes in its neighbors.
OJ‘s undirected graph serialization (so you can understand error output):
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 / \_/
Note: The information about the tree serialization is only meant so that you can understand error output if you get a wrong answer. You don‘t need to understand the serialization to solve the problem.
解题思路:可以采用深度优先搜索,也可以采用广度优先搜索。这里用广度优先搜索进行遍历,用一个map标记访问过的点,并用一个队列辅助遍历。
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) { if(node == nullptr) return nullptr; unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> map; queue<UndirectedGraphNode*> que; que.push(node); map[node] = new UndirectedGraphNode(node->label); while(!que.empty()){ UndirectedGraphNode* t = que.front(); que.pop(); for(auto neighbor:t->neighbors){ if(map.find(neighbor)==map.end()){ map[neighbor] = new UndirectedGraphNode(neighbor->label); que.push(neighbor); } map[t]->neighbors.push_back(map[neighbor]); } } return map[node]; }
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