leetcode -- Clone Graph

不要晃荡，找准方向

[问题描述]

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 #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

 

Visually, the graph looks like the following:

       1
      / \
     /   \
    0 --- 2
         / \
         \_/

[解题思路]
深度优先遍历

UndirectedGraphNode *Solution::cloneGraph(UndirectedGraphNode *node){
    if (node == NULL)
        return NULL;
    map<UndirectedGraphNode*, UndirectedGraphNode*> visited;
    queue<UndirectedGraphNode*> bfs_q;
    visited[node] = new UndirectedGraphNode(node->label);
    bfs_q.push(node);
    while(!bfs_q.empty()){
        UndirectedGraphNode* tmp = bfs_q.front(); bfs_q.pop();
        for (auto k : tmp->neighbors){
            if (visited.find(k) == visited.end()){
                UndirectedGraphNode* t = new UndirectedGraphNode(k->label);
                visited[k] = t;
                visited[tmp]->neighbors.push_back(t);
                bfs_q.push(k);
            }
            else{
                visited[tmp]->neighbors.push_back(visited[k]);
            }
        }
    }
    return visited[node];
}