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：

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> neighbors;

    public Node() {}

    public Node(int _val,List<Node> _neighbors) {
        val = _val;
        neighbors = _neighbors;
    }
};
*/
class Solution {
    public Node cloneGraph(Node node) {
        Map<Node, Node> map = new HashMap<>();
        Queue<Node> queue = new LinkedList<>();
        Node copy = new Node(node.val, new ArrayList<Node>());
        map.put(node, copy);
        queue.offer(node);
        while (!queue.isEmpty()) {
            Node cur = queue.poll();
            for (Node nei : cur.neighbors) {
                if (!map.containsKey(nei)) {
                    map.put(nei, new Node(nei.val, new ArrayList<Node>()));
                    queue.offer(nei);
                }
                map.get(cur).neighbors.add(map.get(nei));
            }
        }
        return map.get(node);
    }
}

网上看了别人的解法：

用Stack写的DFS方法：

public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
    if(node == null)
        return null;
    LinkedList<UndirectedGraphNode> stack = new LinkedList<UndirectedGraphNode>();
    HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
    stack.push(node);
    UndirectedGraphNode copy = new UndirectedGraphNode(node.label);
    map.put(node,copy);
    while(!stack.isEmpty())
    {
        UndirectedGraphNode cur = stack.pop();
        for(int i=0;i<cur.neighbors.size();i++)
        {
            if(!map.containsKey(cur.neighbors.get(i)))
            {
                copy = new UndirectedGraphNode(cur.neighbors.get(i).label);
                map.put(cur.neighbors.get(i),copy);
                stack.push(cur.neighbors.get(i));
            }
            map.get(cur).neighbors.add(map.get(cur.neighbors.get(i)));
        }
    }
    return map.get(node);
}

用Recursion写的DFS方法：

public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
    if(node == null)
        return null;
    HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
    UndirectedGraphNode copy = new UndirectedGraphNode(node.label);
    map.put(node,copy);
    helper(node,map);
    return copy;
}
private void helper(UndirectedGraphNode node, HashMap<UndirectedGraphNode, UndirectedGraphNode> map)
{
    for(int i=0;i<node.neighbors.size();i++)
    { 
        UndirectedGraphNode cur = node.neighbors.get(i);
        if(!map.containsKey(cur))
        {
            UndirectedGraphNode copy = new UndirectedGraphNode(cur.label);
            map.put(cur,copy);
            helper(cur,map);
        }
        map.get(node).neighbors.add(map.get(cur));
    }
}