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
/ \_/
给出一个无向连通图，要求复制
基本思路：
对图的遍历，采取广度优先或者深度优先。
遍历时，需要记住已访问的结点。避免重复访问。这功能可以和下面的map重用。
另外需要一个map， 映射，当前节点，和其对应的复制节点。
访问每一个节点时，需要复制其邻接边。对题目来讲，就是复制其 neighbours数组。
当边所引用的节点不存在时，需要创建此结点。
以下深度优先实现方式。在leetcode上实际执行时间为 72ms。
/**
* 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 node;
stack<UndirectedGraphNode *> s;
unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> m;
s.push(node);
auto root = new UndirectedGraphNode(node->label);
m[node] = root;
while (!s.empty()) {
node = s.top();
s.pop();
auto node_copy = m[node];
for (auto neighbor: node->neighbors) {
auto © = m[neighbor];
if (!copy) {
s.push(neighbor);
copy = new UndirectedGraphNode(neighbor->label);
}
node_copy->neighbors.push_back(copy);
}
}
return root;
}
};即将上面算法的stack换成了queue。
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
if (!node) return node;
queue<UndirectedGraphNode *> q;
unordered_map<UndirectedGraphNode *, UndirectedGraphNode *> m;
q.push(node);
auto root_copy = new UndirectedGraphNode(node->label);
m[node] = root_copy;
while (!q.empty()) {
node = q.front();
q.pop();
auto node_copy = m[node];
for (auto neighbor: node->neighbors) {
auto © = m[neighbor];
if (!copy) {
q.push(neighbor);
copy = new UndirectedGraphNode(neighbor->label);
}
node_copy->neighbors.push_back(copy);
}
}
return root_copy;
}
};原文：http://blog.csdn.net/elton_xiao/article/details/46127467