Clone Graph问题及解法

问题描述：

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/ \\_/

问题分析：

典型的DFS问题。不多说明，直接上代码。

过程详见代码：

/*** 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 == NULL) return NULL;UndirectedGraphNode * newnode = new UndirectedGraphNode(node->label);unordered_map<int, UndirectedGraphNode *> labelmap;labelmap[node->label] = newnode;bl(node, newnode, labelmap);return newnode;}void bl(UndirectedGraphNode * node, UndirectedGraphNode * newnode, unordered_map<int, UndirectedGraphNode *>& labelmap){if (node == NULL) return;for (int i = 0; i < node->neighbors.size(); i++){int label = node->neighbors[i]->label;if (labelmap.count(label))newnode->neighbors.emplace_back(labelmap[label]);else{UndirectedGraphNode * n = new UndirectedGraphNode(label);newnode->neighbors.emplace_back(n);labelmap[label] = n;bl(node->neighbors[i], n, labelmap);}}}
};

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