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。