LeetCode | Clone Graph - c++编程基础

题目

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:

分析

这题与Copy with Random Pointer思路一致，都是通过纪录原节点和新节点的映射关系来构建新的数据结构，只要正确遍历完所有节点即可，如下代码中采用BFS进行遍历。

代码

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.Stack;

public class CloneGraph {
	class UndirectedGraphNode {
		int label;
		ArrayList
    
      neighbors;

		UndirectedGraphNode(int x) {
			label = x;
			neighbors = new ArrayList
     
      (); } } public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) { if (node == null) { return null; } Stack
      
        stack = new Stack
       
        (); Map
        
          map = new HashMap
         
          (); UndirectedGraphNode root = new UndirectedGraphNode(node.label); map.put(node.label, root); stack.push(node); while (!stack.isEmpty()) { UndirectedGraphNode v = stack.pop(); UndirectedGraphNode newV = map.get(v.label); ArrayList
          
            newNeighbors = new ArrayList
           
            (); for (UndirectedGraphNode w : v.neighbors) { UndirectedGraphNode newW = null; if (map.containsKey(w.label)) { newW = map.get(w.label); } else { newW = new UndirectedGraphNode(w.label); map.put(w.label, newW); stack.push(w); } newNeighbors.add(newW); } newV.neighbors = newNeighbors; } return root; } }