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
/ \_/
/**
* 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==NULL)return NULL;
queue<UndirectedGraphNode*> qe;
queue<UndirectedGraphNode*> qe_copy;
map<int, UndirectedGraphNode*> createdMap; //已经复制的节点
//创建第一个节点
//注意本题使用(UndirectedGraphNode*)malloc(sizeof(UndirectedGraphNode))会RUNTIME ERROR
UndirectedGraphNode * newFirstNode = new UndirectedGraphNode(node->label);
qe.push(node);
qe_copy.push(newFirstNode);
createdMap[node->label]=newFirstNode;
while(!qe.empty()){
UndirectedGraphNode * qe_top = qe.front(); qe.pop();
UndirectedGraphNode * qe_copy_top = qe_copy.front();qe_copy.pop();
//创建邻接节点
int neighborsSize = qe_top->neighbors.size();
for(int i=0; i<neighborsSize; i++){
node = qe_top->neighbors[i];
//判断当前节点是否已经创建过
if(createdMap[node->label]!=NULL){
//如果已经创建了,直接从createdmap中取即可
qe_copy_top->neighbors.push_back(createdMap[node->label]);
}
else{
//如果还没创建,则创建新节点
//注意本题使用(UndirectedGraphNode*)malloc(sizeof(UndirectedGraphNode))会RUNTIME ERROR
UndirectedGraphNode * node_copy = new UndirectedGraphNode(node->label);
qe.push(node);
qe_copy.push(node_copy);
createdMap[node->label]=node_copy;
//添加到qe_copy_top的邻接节点
qe_copy_top->neighbors.push_back(node_copy);
}
}
}
return newFirstNode;
}
};LeetCode: Clone Graph [133],布布扣,bubuko.com
原文地址:http://blog.csdn.net/harryhuang1990/article/details/34532037