[各种面试题] 复制有向图

复制一个有向图。输入是有向图中的一个结点指针，返回复制的图中对应的结点指针。有向图中结点的定义为：

C++

struct GraphNode {
    int data;
    vector<GraphNode*> neighbors;
    GraphNode(int data) : data(data) {}
};

跟复制带随即指针的链表其实是一样的。思路就是先复制新节点，然后复制连接关系，最后再把新复制的节点弹出来。

算法一次写出来的，很高兴~ 不过好多笔误。

void createClone(GraphNode* node,set<GraphNode*>& visited);
void cloneNeighbors(GraphNode* node,set<GraphNode*>& visited);
void removeClone(GraphNode* node,set<GraphNode*>& visited);
GraphNode *cloneGraph(GraphNode *node) {
	if ( !node )
		return node;
	set<GraphNode*> visited;
	createClone(node,visited);
	visited.clear();
	cloneNeighbors(node,visited);
	visited.clear();
	GraphNode* newNode = node->neighbors.back();
	removeClone(node,visited);
	return newNode;
}
void createClone(GraphNode* node,set<GraphNode*>& visited)
{
	set<GraphNode*>::iterator it = visited.find(node);
	if ( it!=visited.end())
		return ;
	visited.insert(it,node);
	vector<GraphNode*>& connect=node->neighbors;	
	for(int i=0;i<connect.size();i++)
		createClone(connect[i],visited);
	GraphNode* clone=new GraphNode(node->data);
	connect.push_back(clone);
}
void cloneNeighbors(GraphNode* node,set<GraphNode*>& visited)
{
	set<GraphNode*>::iterator it = visited.find(node);
	if ( it!=visited.end())
		return ;
	visited.insert(it,node);
	vector<GraphNode*>& connect=node->neighbors;	
	vector<GraphNode*>& toAdd=connect.back()->neighbors;
	for(int i=0;i<connect.size()-1;i++)
		cloneNeighbors(connect[i],visited);
	for(int i=0;i<connect.size()-1;i++)
		toAdd.push_back(connect[i]->neighbors.back());
}
void removeClone(GraphNode* node,set<GraphNode*>& visited)
{
	set<GraphNode*>::iterator it = visited.find(node);
	if ( it!=visited.end())
		return ;
	visited.insert(it,node);
	vector<GraphNode*>& connect=node->neighbors;	
	for(int i=0;i<connect.size()-1;i++)
		removeClone(connect[i],visited);
	connect.pop_back();
}