Clone Graph

Clone an undirected graph（无向图）. 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.//感觉这一段没什么用，我在这里卡了很久。
Visually, the graph looks like the following:

   1  / \ /   \0 --- 2     / \     \_/

解题思路：这个题目其实就是无向图的拷贝，关于图我们知道有两种遍历方法，分别为DFS（深度优先）和BFS（宽度优先）。两者没有优劣之说，通常DFS使用栈或者递归来处理，而BFS通常使用队列来解决问题。因为之前做的和图相关的题目大部分都是用的BFS，所以这里我们就看看BFS遍历的实现吧（java代码）：

//定义一个类用于表示图中的节点/* publice class UndirectedGraphNode{        int label;        ArrayList<UndirectedGraphNode> neighbors;        UndirectedGraphNode(int x){            label = x;            neighbors = new ArrayList<UndirectedGraphNode> ();        }    }*/private static void bfs(UndirectedGraphNode node){    if(node == null)        return;    //声明一个链表（相当于队列）来存储下一个遍历的节点    LinkedList<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode> ();    //使用HashMap来存储已经遍历过的节点（Interger表示此时的节点离顶点节点的距离）    HashMap<UndirectedGraphNode,Interger> map = new HashMap<UndirectedGraphNode> ();    //将顶点节点入队    queue.add(node);    map.put(node,0);    int order = 0;    //当队列不为空的时候循环遍历    while(! queue.isEmpty()){        //弹出第一个节点        UndirectedGraphNode curr = queue.pop();        //邻居节点        ArrayList<UndirectedGraphNode> currNeighbors = curr.neighbors;        order ++;        //遍历打印出节点        System.out.printIn("The"+order+"th element"+curr.label+" Distance from the top element is:"+map.get(curr));        int distance = map.get(curr) + 1;        //遍历邻居节点        for(UndirectedGraphNode neighbor : currNeighbors){            //如果该邻居节点还没遍历            if(! map.containsKey(neighbor)){                //将遍历的邻居节点加入到map中                map.put(neighbor,distance);                //将节点加入队列中，下一次再遍历该节点的邻居节点                queue.add(neighbor);            }        }    }}

其实拷贝的过程就是每个节点都遍历一遍，然后将每个节点拷贝一遍。代码如下（java）：

public class Solution {    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {        if(node == null)            return null;        LinkedList<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();        HashMap<UndirectedGraphNode, UndirectedGraphNode> map =                                    new HashMap<UndirectedGraphNode,UndirectedGraphNode>();        UndirectedGraphNode newHead = new UndirectedGraphNode(node.label);        queue.add(node);        map.put(node, newHead);        while(!queue.isEmpty()){            UndirectedGraphNode curr = queue.pop();            ArrayList<UndirectedGraphNode> currNeighbors = curr.neighbors;             for(UndirectedGraphNode aNeighbor: currNeighbors){                if(!map.containsKey(aNeighbor)){                    UndirectedGraphNode copy = new UndirectedGraphNode(aNeighbor.label);                    map.put(aNeighbor,copy);                    map.get(curr).neighbors.add(copy);                    queue.add(aNeighbor);                }else{                    map.get(curr).neighbors.add(map.get(aNeighbor));                }            }        }        return newHead;    }}

代码是不是和遍历的代码差不多？通过这个题目我发现其实很多的题目都是建立在基础知识上的，有的甚至是基础的变形。掌握好基础，多刷刷题目，多做做项目，相信会成长不少。