Leetcode: Clone Graph

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 #.

First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
Second node is labeled as 1. Connect node 1 to node 2.
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
         / \
         \_/

Leetcode里关于图的题其实并不多，这道题就是其中之一。DFS深度优先搜索和BFS广度优先搜索都可以做，遍历完原图的所有节点。这道题的难点在于neighbour关系的拷贝：原图中某一个点跟一些点具有neighbour关系，那么该点的拷贝也要与上述那些点的拷贝具有neighbour关系。那么，就需要很灵活地通过一个点访问该点的拷贝，最好的办法就是把该点与该点的拷贝存入一个HashMap。这样做还有一个好处，就是帮我们剩下了一个visited数组，我们可以用这个HashMap来知道哪些点我是访问过的。

方法是用BFS做的：

 1 /*
 2 // Definition for a Node.
 3 class Node {
 4     public int val;
 5     public List<Node> neighbors;
 6 
 7     public Node() {}
 8 
 9     public Node(int _val,List<Node> _neighbors) {
10         val = _val;
11         neighbors = _neighbors;
12     }
13 };
14 */
15 class Solution {
16     public Node cloneGraph(Node node) {
17         Map<Node, Node> map = new HashMap<>();
18         Queue<Node> queue = new LinkedList<>();
19         Node copy = new Node(node.val, new ArrayList<Node>());
20         map.put(node, copy);
21         queue.offer(node);
22         while (!queue.isEmpty()) {
23             Node cur = queue.poll();
24             for (Node nei : cur.neighbors) {
25                 if (!map.containsKey(nei)) {
26                     map.put(nei, new Node(nei.val, new ArrayList<Node>()));
27                     queue.offer(nei);
28                 }
29                 map.get(cur).neighbors.add(map.get(nei));
30             }
31         }
32         return map.get(node);
33     }
34 }

网上看了别人的解法：

用Stack写的DFS方法：

 1 public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
 2     if(node == null)
 3         return null;
 4     LinkedList<UndirectedGraphNode> stack = new LinkedList<UndirectedGraphNode>();
 5     HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
 6     stack.push(node);
 7     UndirectedGraphNode copy = new UndirectedGraphNode(node.label);
 8     map.put(node,copy);
 9     while(!stack.isEmpty())
10     {
11         UndirectedGraphNode cur = stack.pop();
12         for(int i=0;i<cur.neighbors.size();i++)
13         {
14             if(!map.containsKey(cur.neighbors.get(i)))
15             {
16                 copy = new UndirectedGraphNode(cur.neighbors.get(i).label);
17                 map.put(cur.neighbors.get(i),copy);
18                 stack.push(cur.neighbors.get(i));
19             }
20             map.get(cur).neighbors.add(map.get(cur.neighbors.get(i)));
21         }
22     }
23     return map.get(node);
24 }

用Recursion写的DFS方法：

 1 public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
 2     if(node == null)
 3         return null;
 4     HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
 5     UndirectedGraphNode copy = new UndirectedGraphNode(node.label);
 6     map.put(node,copy);
 7     helper(node,map);
 8     return copy;
 9 }
10 private void helper(UndirectedGraphNode node, HashMap<UndirectedGraphNode, UndirectedGraphNode> map)
11 {
12     for(int i=0;i<node.neighbors.size();i++)
13     { 
14         UndirectedGraphNode cur = node.neighbors.get(i);
15         if(!map.containsKey(cur))
16         {
17             UndirectedGraphNode copy = new UndirectedGraphNode(cur.label);
18             map.put(cur,copy);
19             helper(cur,map);
20         }
21         map.get(node).neighbors.add(map.get(cur));
22     }
23 }

posted @ 2014-10-14 07:08 neverlandly 阅读(411) 评论(0) 收藏举报

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