标签：end 解题思路 uniq 广度优先 pop order wrong 深度优先搜索 like
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.
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 #.
0. Connect node 0 to both nodes 1 and 2.1. Connect node 1 to node 2.2. Connect node 2 to node 2 (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]; }
标签：end 解题思路 uniq 广度优先 pop order wrong 深度优先搜索 like
原文地址：https://www.cnblogs.com/nobodywang/p/10353484.html
