Google+ Followers

Wednesday, December 4, 2013

[LeetCode] Clone Graph, Solution

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
/ \
\_/

 


[Thoughts]


这题和链表拷贝类似:http://fisherlei.blogspot.com/2013/11/leetcode-copy-list-with-random-pointer.html


所不同的是,在链表拷贝中,没有借助额外空间,通过多次链表遍历来拷贝、链接及拆分。


而这里图的拷贝,也可以通过多次遍历来插入拷贝节点,链接拷贝节点以及将拷贝节点拆分出来。但是同样的问题是,需要对图进行多次遍历。如果想在一次遍历中,完成拷贝的话,那就需要使用额外的内存来使用map存储源节点和拷贝节点之间的对应关系。有了这个关系之后,在遍历图的过程中,就可以同时处理访问节点及访问节点的拷贝节点,一次完成。详细看下面代码。


 


[Code]


1 /**
2 * Definition for undirected graph.
3 * struct UndirectedGraphNode {
4 * int label;
5 * vector<UndirectedGraphNode *> neighbors;
6 * UndirectedGraphNode(int x) : label(x) {};
7 * };
8 */
9 class Solution {
10 public:
11 UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
12 if(node == NULL) return NULL;
13 unordered_map<UndirectedGraphNode *, UndirectedGraphNode *> nodeMap;
14 queue<UndirectedGraphNode *> visit;
15 visit.push(node);
16 UndirectedGraphNode * nodeCopy = new UndirectedGraphNode(node->label);
17 nodeMap[node] = nodeCopy;
18 while (visit.size()>0)
19 {
20 UndirectedGraphNode * cur = visit.front();
21 visit.pop();
22 for (int i = 0; i< cur->neighbors.size(); ++i)
23 {
24 UndirectedGraphNode * neighb = cur->neighbors[i];
25 if (nodeMap.find(neighb) == nodeMap.end())
26 {
27 // no copy of neighbor node yet. create one and associate with the copy of cur
28 UndirectedGraphNode* neighbCopy = new UndirectedGraphNode(neighb->label);
29 nodeMap[cur]->neighbors.push_back(neighbCopy);
30 nodeMap[neighb] = neighbCopy;
31 visit.push(neighb);
32 }
33 else
34 {
35 // already a copy there. Associate it with the copy of cur
36 nodeMap[cur]->neighbors.push_back(nodeMap[neighb]);
37 }
38 }
39 }
40
41 return nodeCopy;
42 }
43 };
Post a Comment