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[leetcode] 133 Clone Graph

时间:2015-01-17 19:34:16      阅读:186      评论:0      收藏:0      [点我收藏+]

标签:leetcode   算法   

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
         /          \_/

代码:

UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {  //c++
        if(node == NULL)
            return NULL;
            
        set<int> visited;
        map<int,UndirectedGraphNode*> myMap;
        queue<UndirectedGraphNode*> myQueue;
        UndirectedGraphNode *head,*p,*temp,*one;
        
        myQueue.push(node);
        bool isFirst = true;
        do
        {
            p = myQueue.front();
            if(isFirst){
                temp = new UndirectedGraphNode(p->label);
                head= temp;
                myMap.insert(make_pair(p->label,temp));
                isFirst = false;
            }
            else temp = myMap[p->label];
            vector<UndirectedGraphNode*> vec;
            for(int i=0; i<(p->neighbors).size(); i++){
                int label =(p->neighbors)[i]->label;
                if(myMap.count(label)!=0)
                    one = myMap[label];
                else 
                {
                    one = new UndirectedGraphNode(label);
                    myMap.insert(make_pair(label,one));
                }
                vec.push_back(one);
                if(visited.count(label)==0)
                    myQueue.push(p->neighbors[i]);
            }
            temp->neighbors = vec;
            myQueue.pop();
            visited.insert(p->label);
        }while(!myQueue.empty());
        return head;
    }

[leetcode] 133 Clone Graph

标签:leetcode   算法   

原文地址:http://blog.csdn.net/chenlei0630/article/details/42807633

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