(M)DFS:133. Clone Graph

每次搜索的时候看这个结点是不是已经被创建，是的话就返回其copy，否则就创建，然后再依次深度遍历其邻居结点并将其加入邻居集合中去

/** * 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) return NULL;          if(hash.count(node))    return hash[node];        hash[node] = new UndirectedGraphNode(node->label);        for(auto val : node->neighbors)            hash[node]->neighbors.push_back(cloneGraph(val));        return hash[node];    }private:      unordered_map<UndirectedGraphNode *, UndirectedGraphNode *> hash;  };

还有一种解法，这道题和前面一个random list的copy的题类似，关键点是建立一个map存放旧数据结构里的节点和新数据结构里的节点的对应关系。对于这个题，我一开始觉得应该先遍历一遍图，每个节点重建，完善map结构。然后再遍历一遍，把每个节点的neighbors填好。但是其实一遍遍历就可以做完：

/** * 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) return NULL;        map<UndirectedGraphNode*, UndirectedGraphNode*> m;        queue<UndirectedGraphNode*> q;        q.push(node);        while(!q.empty()){            UndirectedGraphNode *n = q.front();            q.pop();            if(!m[n])            {            UndirectedGraphNode *newnode = new UndirectedGraphNode(n->label);                m[n] = newnode;            }                        for(auto a : n->neighbors){                if(!m[a])                {                    UndirectedGraphNode *newa = new UndirectedGraphNode(a->label);                    m[a] = newa;                    q.push(a);                }                m[n]->neighbors.push_back(m[a]);            }        }        return m[node];    }};