Clone Graph Leetcode 133

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     / \     \_/

题目的要求是让我们实现一个图的拷贝。拷贝指的是复制另外一个对象到自己的对象中，且两者不共享一个内存区，所以我们需要new 来开辟新的内存区执行拷贝。

analysis：
在一次遍历中，完成拷贝的话，那就需要使用额外的内存来使用map存储源节点和拷贝节点之间的对应关系。有了这个关系之后，在遍历图的过程中，就可以同时处理访问节点及访问节点的拷贝节点，一次完成。

1.用unordered_map(node*,node*)这种结构，直接map[node]=nodeclone,较为简便。
2.BFS，变搜边向map添加元素.
3.用queue来存储要copy的节点

/** * 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;        UndirectedGraphNode *copy=new UndirectedGraphNode(node->label);        map<UndirectedGraphNode *,UndirectedGraphNode *>table;        table[node]=copy;        queue<UndirectedGraphNode *>visit;        visit.push(node);        while(!visit.empty()){            UndirectedGraphNode *cur=visit.front();            visit.pop();            for(int i=0;i<cur->neighbors.size();i++){                UndirectedGraphNode *neighbor=cur->neighbors[i];                if(table.find(neighbor)==table.end()){                    UndirectedGraphNode *neighborclone=new UndirectedGraphNode(neighbor->label);                    table[neighbor]=neighborclone;                    visit.push(neighbor);                }                table[cur]->neighbors.push_back(table[neighbor]);            }        }        return copy;    }};

solution 2

class Solution {  public:      UndirectedGraphNode *clone(UndirectedGraphNode *node, map<int,UndirectedGraphNode *> &table) {        if(node==NULL)return NULL;        if(table.find(node->label)!=table.end())return table[node->label];        UndirectedGraphNode *copy=new UndirectedGraphNode(node->label);        table[copy->label]=copy;        for(int i=0;i<node->neighbors.size();i++){            UndirectedGraphNode *neighborclone=clone(node->neighbor[i],table);            copy->neighbors.push_back(neighborclone);        }        return copy;    }    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {        map<int,UndirectedGraphNode *> visit;        return clone(node,visit);    }};