LeetCode 133. Clone Graph (dfs,複製圖)

Given the head of a graph, return a deep copy (clone) of the graph. Each node in the graph contains a label (int) and a list (List[UndirectedGraphNode]) of its neighbors. There is an edge between the given node and each of the nodes in its neighbors.

OJ’s undirected graph serialization (so you can understand error output):

Nodes are labeled uniquely.

#
,

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
         / \
         \_/

Note: The information about the tree serialization is only meant so that you can understand error output if you get a wrong answer. You don’t need to understand the serialization to solve the problem.

解法
使用一個map記錄節點的值和節點指針。使用dfs深度優先搜索圖,每次創建節點的時候就在map裏面增加一條記錄。
(注意:需要在一開始的時候就增加map記錄,否則會出現死循環的問題,例如

{0,0,0}

這個圖,如果等到找到節點的時候再增加記錄則出錯。)

錯誤代碼

class Solution {
    map<int,  UndirectedGraphNode *> dict;
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(node == nullptr)
            return node;
        UndirectedGraphNode *root =new  UndirectedGraphNode(node->label);
        for(int i=0;i<node->neighbors.size();i++) {
            if(dict.find(node->neighbors[i]->label)==dict.end())
                 dict[node->neighbors[i]->label] = cloneGraph(node->neighbors[i]); //在這才記錄可能死循環
            root->neighbors.push_back(dict[node->neighbors[i]->label]);
        }
        return root;
    }
};

正確代碼

class Solution {
    map<int,  UndirectedGraphNode *> dict;
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(node == nullptr)
            return node;
        UndirectedGraphNode *root =new  UndirectedGraphNode(node->label);
        dict[root->label] = root; // 在這裏記錄數據
        for(int i=0;i<node->neighbors.size();i++) {
            if(dict.find(node->neighbors[i]->label)==dict.end())
               cloneGraph(node->neighbors[i]);
            root->neighbors.push_back(dict[node->neighbors[i]->label]);
        }
        return root;
    }
};
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