【LeetCode】#133克隆圖(Clone Graph)

題目描述

克隆一張無向圖，圖中的每個節點包含一個 label （標籤）和一個 neighbors （鄰接點）列表 。
OJ的無向圖序列化：
節點被唯一標記。
我們用 # 作為每個節點的分隔符，用 , 作為節點標籤和鄰接點的分隔符。

示例

例如，序列化無向圖 {0,1,2#1,2#2,2}。
該圖總共有三個節點, 被兩個分隔符 # 分為三部分。
第一個節點的標籤為 0，存在從節點 0 到節點 1 和節點 2 的兩條邊。
第二個節點的標籤為 1，存在從節點 1 到節點 2 的一條邊。
第三個節點的標籤為 2，存在從節點 2 到節點 2 (本身) 的一條邊，從而形成自環。
我們將圖形視覺化如下：

   1
  / \
 /   \
0 --- 2
     / \
     \_/

Description

Given the head of a graph, return a deep copy (clone) of the graph. Each node in the graph contains a label (int) and a list (List[UndirectedGraphNode]) of its neighbors. There is an edge between the given node and each of the nodes in its neighbors.
OJ’s undirected graph serialization (so you can understand error output):
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.

Example

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
     / \
     \_/

解法

public class Solution {
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if (node == null) {
            return node;
        }
        UndirectedGraphNode newRoot = new UndirectedGraphNode(node.label);
        Map<Integer, UndirectedGraphNode> done = new HashMap<>();
        done.put(node.label, newRoot);
        if(node.neighbors != null && !node.neighbors.isEmpty()) {
            process(newRoot, node.neighbors, done);
        }
        return newRoot;
    }
    
    private void process(UndirectedGraphNode node, List<UndirectedGraphNode> neighbors,  Map<Integer, UndirectedGraphNode> done) {
        for(UndirectedGraphNode neighbor : neighbors) {
            UndirectedGraphNode newNeighbor;
            if (!done.containsKey(neighbor.label)) {
                newNeighbor= new UndirectedGraphNode(neighbor.label);  
                done.put(newNeighbor.label, newNeighbor);
                if(neighbor.neighbors != null && !neighbor.neighbors.isEmpty()) {
                    process(newNeighbor, neighbor.neighbors, done);
                }
            } else {
                newNeighbor = done.get(neighbor.label);
            }
            node.neighbors.add(newNeighbor);
        }
    }
}