133. Clone Graph

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题目描述

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:

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

题意分析

克隆一个图
每次添加一个节点, 并且保存他的邻居信息.

其中一个难点就是邻居可能已经出现过，你只要把他的指针加到邻居集合中即可，也有可能这个结点还没出现过，因此你需要新建一个这个结点，因此我们需要一个hash表来对结点做一一映射．

本题有两种方法来做，广度搜索BFS和深度搜索DFS.
我选择了DFS,更简洁.
参考了别人的更优雅的代码, 改进了自己的代码

实现

/** * Definition for undirected graph. * struct UndirectedGraphNode { *     int label; *     vector<UndirectedGraphNode *> neighbors; *     UndirectedGraphNode(int x) : label(x) {}; * }; */class Solution {  public:      UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {          if(!node) return NULL;  // empty node         if(hash.count(node)) return hash[node]; // the node exists        hash[node] = new UndirectedGraphNode(node->label);  // create the node        for(auto neighbor: node->neighbors)   // add neighbors            hash[node]->neighbors.push_back(cloneGraph(neighbor));          return hash[node];      }  private:      unordered_map<UndirectedGraphNode *, UndirectedGraphNode *> hash;  };