[leetcode-133]Clone Graph（java）

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 #.

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:

[code]   1
  / \
 /   \
0 --- 2
     / \
     \_/

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分析：这道题的意思是完整复制一份图，实际上考察的是图的遍历方法，这道题主要使用BFS，然后先复制点，再复制边。旧图与新图之间使用一个map来做映射。而且新节点中的neibour都是要新创建出来的。

代码如下：388ms

[code]/**
 * Definition for undirected graph.
 * 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 null;
        UndirectedGraphNode newNode = new UndirectedGraphNode(node.label);
        HashMap<UndirectedGraphNode,UndirectedGraphNode> maps = new HashMap<>();
        Queue<UndirectedGraphNode> queue = new LinkedList<>();

        maps.put(node,newNode);
        queue.offer(node);

        while (!queue.isEmpty()){
            UndirectedGraphNode top = queue.poll();
            List<UndirectedGraphNode> neighbors = top.neighbors;

            for(UndirectedGraphNode neighbour:neighbors){
                if(!maps.containsKey(neighbour)){
                    UndirectedGraphNode newTmpNode = new UndirectedGraphNode(neighbour.label);
                    maps.put(neighbour,newTmpNode);
                    queue.offer(neighbour);
                }
                maps.get(top).neighbors.add(maps.get(neighbour));
            }
        }
        return newNode;
    }
}