[LeetCode]: 133: 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
         / \
         \_/

 

题目分析:

题目本身就是个图的遍历问题,其实只用看最上面的一句就OK了。下面整这么一篇就是为了说明Testcase,结果误导对了我好久。

与树的遍历不同之处就是图可能会出现"闭合的环"这样会造成无线循环,所以遍历的时候需要把已经遍历过的点去掉。

根据题目,图中每一个点的权值是不一样的,所以可以采用取巧的方法检测这个点是否已经加在了图中,即使用HashMap

 

思路一:深度优先搜索

    public static UndirectedGraphNode cloneDFS(UndirectedGraphNode node,HashMap<Integer, UndirectedGraphNode> validator){
        if(validator.containsKey(node.label)){
            return validator.get(node.label);
        }
        
        UndirectedGraphNode nodeTemp = new UndirectedGraphNode(node.label);
        validator.put(node.label, nodeTemp);
        
        for(int i =0;i<node.neighbors.size();i++){
            nodeTemp.neighbors.add(cloneDFS(node.neighbors.get(i),validator));
        }
        
        return nodeTemp;
    }
    public static UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if(node== null ||  node.equals(null)){
            return null;
        }
        
        HashMap<Integer, UndirectedGraphNode> validator = new HashMap<Integer, UndirectedGraphNode>();
        return cloneDFS(node, validator);
    }

 

思路二:广度优先搜索

XXX

 
 
posted @ 2015-10-16 11:50  savageclc26  阅读(186)  评论(0编辑  收藏  举报