LeetCode Clone Graph

LeetCode Clone Graph

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
         / \
         \_/

给定一个无向图,要求克隆一个一样的图,也就是深拷贝。

图的问题,想到应该是DFS或者BFS,但是没太想明白,因为DFS过程中可能遇到之前已经new过的节点,不知道怎么判断比较好,查看讨论区,发现每new一个节点,把这个节点存到一个hash表中就好了,下回先查hash表,不中才new新的。

因为每个节点的label是unique的,所以hash的key可以是label,value是node*。

完整代码如下,注意第9行,new出来的节点要先加入hash,再递归DFS,否则递归DFS在hash表中找不到当前的节点,比如{0,0,0}这个样例就会出错。

class Solution {
private:
	unordered_map<int, UndirectedGraphNode*> graph;
public:
	UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
		if (node == NULL)return NULL;
		if (graph.find(node->label) != graph.end())return graph[node->label];
		UndirectedGraphNode *cur = new UndirectedGraphNode(node->label);
		graph[cur->label] = cur;
		for (auto& n : node->neighbors) {
			cur->neighbors.push_back(cloneGraph(n));
		}
		return cur;
	}
};

本代码提交AC,用时26MS。

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