LeetCode | 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 #.

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         / \         \_/

思路：

首先做一次BFS遍历，将所有结点创建出来。并且要保存在一个可以在短时间读取的结构中。map是个不错的结构并且每个结点label唯一，因此我们可以用label来作为索引。

第二次遍历的时候可以快速地找出结点并构造图的关系。

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;        }        map<int, UndirectedGraphNode*> nodes;        queue<UndirectedGraphNode*> q;                q.push(node);                while(!q.empty())        {            UndirectedGraphNode* tmp = q.front();            q.pop();            if(nodes.find(tmp->label)==nodes.end())            {                UndirectedGraphNode* new_node = new UndirectedGraphNode(tmp->label);                nodes.insert(pair<int, UndirectedGraphNode*>(new_node->label, new_node));                for(int i=0;i<tmp->neighbors.size();i++)                {                    q.push(tmp->neighbors[i]);                }            }        }                q.push(node);        while(!q.empty())        {            UndirectedGraphNode* tmp = q.front();            q.pop();            UndirectedGraphNode* existingnode = nodes[tmp->label];            if(existingnode->neighbors.empty()&&!tmp->neighbors.empty())            {                for(int i=0;i<tmp->neighbors.size();i++)                {                    existingnode->neighbors.push_back(nodes[tmp->neighbors[i]->label]);                    q.push(tmp->neighbors[i]);                }            }        }                return nodes[node->label];    }};