[LeetCode]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 #.

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
         / \
         \_/

BFS，用hash来存储新的图和旧图的一一对应关系的点，采用BFS搜索。然后对每一个点存相邻点。

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        unordered_map<UndirectedGraphNode * ,UndirectedGraphNode *> NodeMap;
        //和克隆图之间的Map这样可以一次遍历完
        queue<UndirectedGraphNode *> Qq;
        if(node == NULL)
            return node;
        Qq.push(node);
        while(!Qq.empty()){
            UndirectedGraphNode *top = Qq.front();
            Qq.pop();
            if(NodeMap.count(top)==0){//没有节点
                UndirectedGraphNode * Temp = new UndirectedGraphNode(top->label);
                NodeMap[top] = Temp;
            }
            //保存相邻节点
            for(auto it = top->neighbors.begin(); it!=top->neighbors.end(); ++it){
                UndirectedGraphNode *adj = *it;
                if(NodeMap.count(adj)==0){//没有该节点，创建节点，同时间也避免了环，之前遍历过的节点不会再遍历
                    UndirectedGraphNode *Temp = new UndirectedGraphNode(adj->label);
                    NodeMap[adj] = Temp;
                    Qq.push(adj);
                }
                NodeMap[top]->neighbors.push_back(NodeMap[adj]); //给定字节点所有链接节点，如果有环也会保留，因为没有入队列
            }
        }
        return NodeMap[node];
    }
};