133-Clone Graph

2024年1月21日 128次阅读

Description

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
         / \          \_/

问题描述

复制一个无向图。图里的每一个节点包含一个标签还有相邻节点列表

在线评测的无向图序列化:

每个节点的标签唯一

使用’#’分隔每个节点， 使用’,’分隔节点标签以及相邻节点

举例来说， 考虑{0, 1, 2#1, 2 # 2, 2}

这个有向图总共有3个节点， 因此被’#’分隔后有3部分

  1. 第一个节点被标记为0.将其与1和2相连
  2. 第二个节点被标记为1.将其与2相连(注意没有0)
  3. 第三个节点被标记为2， 将其与2相连， 形成自环

有向图如下所示：

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

问题分析

遍历节点， 创建新节点即可， 注意用Map保存已经遇到过的节点

解法

public class Solution {
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        HashMap<Integer,UndirectedGraphNode> map = new HashMap();

        return dfs(node,map);
    }
    private UndirectedGraphNode dfs(UndirectedGraphNode node, HashMap<Integer,UndirectedGraphNode> map) {
        if(node == null) return null;
        if(map.containsKey(node.label)) return map.get(node.label);

        UndirectedGraphNode clone = new UndirectedGraphNode(node.label);
        map.put(node.label,clone);
        for(int i = 0; i < node.neighbors.size(); i++) clone.neighbors.add(dfs(node.neighbors.get(i), map));

        return clone;
    }
}
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