LEETCODE，leetcode -- Clone Graph

2023-10-07 阅读 29 评论 0

摘要：Clone an undirected graph. Each node in the graph contains a?label?and a list of its?neighbors. [解題思路] 圖的遍歷有兩種方式，BFS和DFS 這里使用BFS來解本題，BFS需要使用queue來保存neighbors LEETCODE。但這里有個問題，在clone一個節點時我們需

Clone an undirected graph. Each node in the graph contains a?label?and a list of its?neighbors.

[解題思路]

圖的遍歷有兩種方式，BFS和DFS

這里使用BFS來解本題，BFS需要使用queue來保存neighbors

LEETCODE。但這里有個問題，在clone一個節點時我們需要clone它的neighbors，而鄰居節點有的已經存在，有的未存在，如何進行區分？

這里我們使用Map來進行區分，Map的key值為原來的node，value為新clone的node，當發現一個node未在map中時說明這個node還未被clone，

將它clone后放入queue中處理neighbors。

使用Map的主要意義在于充當BFS中Visited數組，它也可以去環問題，例如A--B有條邊，當處理完A的鄰居node，然后處理B節點鄰居node時發現A已經處理過了

處理就結束，不會出現死循環！

leetcode 5，queue中放置的節點都是未處理neighbors的節點！！！！

 1 /**
 2  * Definition for undirected graph.
 3  * class UndirectedGraphNode {
 4  *     int label;
 5  *     ArrayList<UndirectedGraphNode> neighbors;
 6  *     UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
 7  * };
 8  */
 9 public class Solution {
10     public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
11         // Note: The Solution object is instantiated only once and is reused by each test case.
12         if(node == null){
13             return node;
14         }
15         UndirectedGraphNode result = new UndirectedGraphNode(node.label);
16         LinkedList<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
17         queue.add(node);
18         Map<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
19         map.put(node, result);
20         
21         while(!queue.isEmpty()){
22             UndirectedGraphNode nodeInQueue = queue.poll();
23             ArrayList<UndirectedGraphNode> neighbors = nodeInQueue.neighbors;
24             for(int i = 0; i < neighbors.size(); i++){
25                 UndirectedGraphNode n1 = neighbors.get(i);
26                 if(map.containsKey(n1)){
27                     map.get(nodeInQueue).neighbors.add(map.get(n1));
28                 } else {
29                     UndirectedGraphNode n1clone = new UndirectedGraphNode(n1.label);
30                     map.get(nodeInQueue).neighbors.add(n1clone);
31                     map.put(n1, n1clone);
32                     queue.add(n1);
33                 }
34             }
35             
36         }
37         return result;
38     }
39 }