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; } };
原文地址:http://blog.csdn.net/elton_xiao/article/details/46127467