leetcode clone graph

题目描述

https://oj.leetcode.com/problems/clone-graph/

思路

第一遍遍历图，在每个节点的neighbors列表的末尾加上一个selfCopyNode，即对该节点的复制节点。

第二遍遍历图，对每个节点的复制节点，让它的每个neighbor都是原节点对应neighbor复制节点。

第三遍遍历图，移除每个节点的selfCopyNode，还原原来的graph，否则无法AC。

代码

class UndirectedGraphNode {int label;List<UndirectedGraphNode> neighbors;UndirectedGraphNode(int x) {label = x;neighbors = new ArrayList<UndirectedGraphNode>();}};public class Solution {public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {if(node==null)return node;Queue<UndirectedGraphNode> q=new LinkedList<UndirectedGraphNode>();Set<Integer> visit=new HashSet<Integer>();visit.add(node.label);q.offer(node);while(!q.isEmpty()){UndirectedGraphNode pNode=q.poll();for(UndirectedGraphNode neighbor:pNode.neighbors){if(!visit.contains(neighbor.label)){q.offer(neighbor);visit.add(neighbor.label);}}UndirectedGraphNode selfCopyNode=new UndirectedGraphNode(pNode.label);pNode.neighbors.add(selfCopyNode);}visit.clear();visit.add(node.label);q.offer(node);while(!q.isEmpty()){UndirectedGraphNode pNode=q.poll();int size=pNode.neighbors.size();UndirectedGraphNode selfCopyNode=pNode.neighbors.get(size-1);for(int i=0;i<size-1;i++){UndirectedGraphNode neighbor=pNode.neighbors.get(i);if(!visit.contains(neighbor.label)){q.offer(neighbor);visit.add(neighbor.label);}selfCopyNode.neighbors.add(neighbor.neighbors.get(neighbor.neighbors.size()-1));}}UndirectedGraphNode root=node.neighbors.get(node.neighbors.size()-1);visit.clear();visit.add(node.label);q.offer(node);while(!q.isEmpty()){UndirectedGraphNode pNode=q.poll();int size=pNode.neighbors.size();pNode.neighbors.remove(size-1);for(int i=0;i<size-1;i++){UndirectedGraphNode neighbor=pNode.neighbors.get(i);if(!visit.contains(neighbor.label)){q.offer(neighbor);visit.add(neighbor.label);}}}return root;}public static void main(String[] args) {Solution sln = new Solution();UndirectedGraphNode node=new UndirectedGraphNode(0);node.neighbors.add(node);node.neighbors.add(node);sln.cloneGraph(node);}}

注意点

处理图为空的情况。