leetcode -- Clone Graph
Clone an undirected graph. Each node in the graph contains a label
and a list of its neighbors
.
[解题思路]
图的遍历有两种方式,BFS和DFS
这里使用BFS来解本题,BFS需要使用queue来保存neighbors
但这里有个问题,在clone一个节点时我们需要clone它的neighbors,而邻居节点有的已经存在,有的未存在,如何进行区分?
这里我们使用Map来进行区分,Map的key值为原来的node,value为新clone的node,当发现一个node未在map中时说明这个node还未被clone,
将它clone后放入queue中处理neighbors。
使用Map的主要意义在于充当BFS中Visited数组,它也可以去环问题,例如A--B有条边,当处理完A的邻居node,然后处理B节点邻居node时发现A已经处理过了
处理就结束,不会出现死循环!
queue中放置的节点都是未处理neighbors的节点!!!!
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 }
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