LeetCode133. Clone Graph

题目：

https://leetcode.com/problems/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 #.

First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
Second node is labeled as 1. Connect node 1 to node 2.
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
         / \          \_/

思路：

题意：克隆一个图。

BFS遍历图，一边遍历，一边完成克隆。用map记录原来图的节点和克隆图的节点情况，防止克隆过的节点再被new一次。每copy一个原节点，就把原节点与克隆节点放入map中，原节点与克隆节点对应。如果后面的遍历遇到在map中的原节点，表明这个原节点已经被copy过了，不用新new一个节点来再次copy。

代码：

/** * 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* cur=NULL;
        UndirectedGraphNode* start=NULL;
        UndirectedGraphNode* connect=NULL;
        queue<UndirectedGraphNode*> q;
        q.push(node);
        unordered_map<UndirectedGraphNode*,UndirectedGraphNode*> m;
        start=new UndirectedGraphNode (node->label);
        m[node]=start;
        while(!q.empty()){
            cur=q.front();
            q.pop();
            int len=cur->neighbors.size();
            for(int i=0;i<len;i++){
                connect=cur->neighbors[i];
                if(m.find(connect)==m.end()){
                    UndirectedGraphNode* nodeCopy=new UndirectedGraphNode(connect->label);
                    m[connect]=nodeCopy;
                    q.push(connect);
                }
                m[cur]->neighbors.push_back(m[connect]);//与原节点对应的克隆节点添加相邻节点
            }
        }
        return start;
    }
};