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Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.
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 #.
0. Connect node 0 to both nodes 1 and 2.1. Connect node 1 to node 2.2. Connect node 2 to node 2 (itself), thus forming a self-cycle.
Visually, the graph looks like the following:
1
/ / 0 --- 2
/ \_/
这道无向图的复制问题和之前的拷贝带有随机指针的链表有些类似，那道题的难点是如何处理每个节点的随机指针，这道题目的难点在于如何处理每个节点的neighbors，由于在深度拷贝每一个节点后，还要将其所有neighbors放到一个vector中，而如何避免重复拷贝呢？这道题好就好在所有节点值不同，所以我们可以使用哈希表来对应节点值和新生成的节点。对于图的遍历的两大基本方法是深度优先搜索DFS和广度优先搜索BFS，此题的两种解法可参见网友爱做饭的小莹子的博客，这里我们使用深度优先搜索DFS来解答此题，代码如下：
/** * Definition for undirected graph. * struct UndirectedGraphNode { * int label; * vector<UndirectedGraphNode *> neighbors; * UndirectedGraphNode(int x) : label(x) {}; * }; */ class Solution { public: UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) { unordered_map<int, UndirectedGraphNode*> umap; return clone(node, umap); } UndirectedGraphNode *clone(UndirectedGraphNode *node, unordered_map<int, UndirectedGraphNode*> &umap) { if (!node) return node; if (umap.count(node->label)) return umap[node->label]; UndirectedGraphNode *newNode = new UndirectedGraphNode(node->label); umap[node->label] = newNode; for (int i = 0; i < node->neighbors.size(); ++i) { (newNode->neighbors).push_back(clone(node->neighbors[i], umap)); } return newNode; } };
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原文地址：http://www.cnblogs.com/grandyang/p/4267628.html
