题目链接：Clone Graph

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 #.

  1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
  2. Second node is labeled as 1. Connect node 1 to node 2.
  3. 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 
         / \ 
         \_/

这道题的要求是克隆一个无向图。

一个小技巧就是用hash表将图中节点与克隆的对应节点关联起来。剩下的就可以用BFS或DFS进行遍历图，当遍历到某一节点时，如果hash表中没有与之对应的克隆的节点，就需要为该节点新建一节点。

  1. BFS

时间复杂度：O(n)

空间复杂度：O(n)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32

class Solution
{
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node)
    {
        if(node == NULL)
            return NULL;
        
        unordered_map<UndirectedGraphNode *, UndirectedGraphNode *> map;
        map[node] = new UndirectedGraphNode(node -> label);
        
        queue<UndirectedGraphNode *> q;
        q.push(node);
        while(!q.empty())
        {
            UndirectedGraphNode *temp = q.front();
            q.pop();
            
            for(int i = 0; i < temp -> neighbors.size(); ++ i)
            {
                if(map.find(temp -> neighbors[i]) == map.end())
                {
                    map[temp -> neighbors[i]] = new UndirectedGraphNode(temp -> neighbors[i] -> label);
                    q.push(temp -> neighbors[i]);
                }
                (map[temp] -> neighbors).push_back(map[temp -> neighbors[i]]);
            }
        }
        
        return map[node];
    }
};

  1. DFS

时间复杂度：O(n)

空间复杂度：O(n)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18

class Solution
{
    unordered_map<UndirectedGraphNode *, UndirectedGraphNode *> map;
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node)
    {
        if(node == NULL)
            return NULL;
        
        if(map.find(node) == map.end())
        {
            map[node] = new UndirectedGraphNode(node -> label);
            for(auto n : node -> neighbors)
                (map[node] -> neighbors).push_back(cloneGraph(n));
        }
        return map[node];
    }
};