• Clone Graph


    问题描述

    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
             / 
             \_/

    解决思路

    因为节点的label值唯一,可以使用一个HashMap来记录<node.label, copy node>。

    使用BFS来遍历所有的原始图上的节点,通过map查找到复制节点,然后添加neighbors。

    程序

    public class Solution {
        public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
            if (node == null) {
                return null;
            }
            
            HashMap<Integer, UndirectedGraphNode> map = new HashMap<>(); // label unique
            Queue<UndirectedGraphNode> q = new LinkedList<>();
            q.offer(node);
            
            while (!q.isEmpty()) {
                // bfs
                int size = q.size();
                for (int i = 0; i < size; i++) {
                    UndirectedGraphNode top = q.poll();
                    if (!map.containsKey(top.label)) {
                        map.put(top.label, new UndirectedGraphNode(top.label));
                    }
                    UndirectedGraphNode copy = map.get(top.label);
                    List<UndirectedGraphNode> neighbors = top.neighbors;
                    
                    if (neighbors != null && neighbors.size() > 0) {
                        for (UndirectedGraphNode n : neighbors) {
                            if (!map.containsKey(n.label)) {
                                map.put(n.label, new UndirectedGraphNode(n.label));
                                q.offer(n);
                            }
                            copy.neighbors.add(map.get(n.label));
                        }
                    }
                }
            }
            
            return map.get(node.label);
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/harrygogo/p/4728245.html
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