Leetcode Clone Graph

Clone Graph

Total Accepted:4360 Total Submissions:21850 My Submissions

Clone an undirected graph. Each node in the graph contains alabeland a list of itsneighbors.

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 as0. Connect node0to both nodes1and2.
2. Second node is labeled as1. Connect node1to node2.
3. Third node is labeled as2. Connect node2to node2(itself), thus forming a self-cycle.

Visually, the graph looks like the following:

       1
      / \
     /   \
    0 --- 2
         / \
         \_/

这里使用深度优先搜索。这样可以递归实现，如果是宽度优先，就要额外使用queue容器。

关键点：

1 这里的clone需要深度拷贝，就是要使用new操作了

2 防止回路无限循环，就要使用hash表，这里使用unordered_map记录访问过的节点。因为这里的label应该是唯一的才对，所以可以直接使用label作为关键字就可以。

看起来挺难的，因为图总给人困难的感觉，其实不难，3到4星级难度吧，很多都是基本操作组合起来。我一次性通过了。

struct UndirectedGraphNode 
{
	int label;
	vector neighbors;
	UndirectedGraphNode(int x) : label(x) {};
};

class Solution 
{
public:
	UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) 
	{
		unordered_map track;
		return cloneGraph(node, track);
	}

	UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node, unordered_map &track) 
	{
		if (!node) return NULL;
		if (track.count(node->label)) return track[node->label];

		UndirectedGraphNode *new_node = new UndirectedGraphNode(node->label);
		new_node->neighbors.resize(node->neighbors.size());
		track[node->label] = new_node;

		for (int i = 0; i < node->neighbors.size(); i++)
		{
			new_node->neighbors[i] = cloneGraph(node->neighbors[i], track);
		}
		return new_node;
	}
};

//2014-2-18 update
	UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) 
	{
		unordered_map ump_iu;
		return clone(node, ump_iu);
	}
	UndirectedGraphNode *clone(UndirectedGraphNode *n, 
		unordered_map &ump_iu)
	{
		if (!n) return n;
		if (ump_iu.count(n->label)) return ump_iu[n->label];

		UndirectedGraphNode *rs = new UndirectedGraphNode(n->label);
		ump_iu[n->label] = rs;
		for (int i = 0; i < n->neighbors.size(); i++)
		{
			(rs->neighbors).push_back(clone((n->neighbors[i]), ump_iu));
		}
		return rs;
	}