leetcode 133 Clone Graph
z
 2019-03-31 13:28:11  

Given a reference of a node in a connected undirected graph, return a deep copy (clone) of the graph. Each node in the graph contains a val (int) and a list (List[Node]) of its neighbors.

Example:

img

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Input:
{"$id":"1","neighbors":[{"$id":"2","neighbors":[{"$ref":"1"},{"$id":"3","neighbors":[{"$ref":"2"},{"$id":"4","neighbors":[{"$ref":"3"},{"$ref":"1"}],"val":4}],"val":3}],"val":2},{"$ref":"4"}],"val":1}

Explanation:
Node 1's value is 1, and it has two neighbors: Node 2 and 4.
Node 2's value is 2, and it has two neighbors: Node 1 and 3.
Node 3's value is 3, and it has two neighbors: Node 2 and 4.
Node 4's value is 4, and it has two neighbors: Node 1 and 3.

Note:

  1. The number of nodes will be between 1 and 100.
  2. The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph.
  3. Since the graph is undirected, if node p has node q as neighbor, then node q must have node p as neighbor too.
  4. You must return the copy of the given node as a reference to the cloned graph.

  • 使用一个map来保存源节点Node和与之对应的deep copy of Node的对应关系。

  • 遍历Node的每一个neighbor， 将如果neighbor没有保存在map中，建立一个neighbor和deep copy of neighbor的对应关系。并将deep copy of neighbor加入deep copy of Node的neighbots中。

  • 对Node的每一个neighbor重复上述操作。

  • 需要注意的是，每次输入的Node有可能已经操作过了，因此，每次操作开始前，需要判断Node是否已经处理过，即判断deep copy of Node 的neighbors长度是否为0即可

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/*
// 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) {
        HashMap<Node, Node> m = new HashMap<>();
        dfs(m, node);
        return m.get(node);
    }
    
    public void dfs(HashMap<Node, Node> m, Node node) {
        if (! m.containsKey(node)) {
            m.put(node, new Node(node.val, new LinkedList<Node>()));
        }
        if ( m.get(node).neighbors.size() != 0)
            return ;
        for (Node n: node.neighbors) {
            if (!m.containsKey(n)) {
                m.put(n, new Node(n.val, new LinkedList<Node>()));
            }
            m.get(node).neighbors.add(m.get(n));
            dfs(m, n);
        }
    }
    
}