【LeetCode】Clone Graph (2 solutions)

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
         /          \_/

这题只需一边遍历一遍复制就可以了。

因此至少可以用三种方法：

1、广度优先遍历(BFS)

2、深度优先遍历(DFS)

2.1、递归

2.2、非递归

解法一：广度优先遍历

变量说明：

hash表m用来保存原图结点与克隆结点的对应关系。

hash表f用来记录已经访问过的原图结点，防止循环访问。

队列q用于记录深度优先遍历的层次信息。

/**
 * 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;

        map<UndirectedGraphNode *, UndirectedGraphNode *> m; //record the <original node, clone node> pairs
        map<UndirectedGraphNode *, bool> f; //record original nodes that have been dealt with
        queue<UndirectedGraphNode *> q; //BFS, q store the original node
        UndirectedGraphNode *nodeClone = new UndirectedGraphNode(node->label);
        m[node] = nodeClone;  //add <node, ret> to map
        q.push(node);
        f[node] = true;

        while(!q.empty())
        {
            UndirectedGraphNode *temp = q.front();  //original node
            UndirectedGraphNode *tempClone = (m.find(temp))->second;   //clone node
            q.pop();
            for(vector<UndirectedGraphNode *>::size_type st = 0; st < temp->neighbors.size(); st ++)
            {//add neighbors of temp to the queue
                UndirectedGraphNode *neighbor = temp->neighbors[st];
                map<UndirectedGraphNode *, bool>::iterator itf = f.find(neighbor);
                if(itf == f.end())
                {
                    f[neighbor] = true;
                    q.push(neighbor);
                }

                map<UndirectedGraphNode *, UndirectedGraphNode *>::iterator it = m.find(neighbor);

                UndirectedGraphNode *neighborClone;
                if(it == m.end())
                {//neighbors[st] not constructed yet
                    neighborClone = new UndirectedGraphNode(neighbor->label);
                    m[neighbor] = neighborClone;
                }
                else
                    neighborClone = it->second;

                //connect neighbor to tempClone
                tempClone->neighbors.push_back(neighborClone);
            }
        }
        return nodeClone;
    }
};

解法二：递归深度优先遍历(DFS)

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    map<UndirectedGraphNode *, UndirectedGraphNode *> f;

    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node)
    {
        if(node == NULL)
            return NULL;

        map<UndirectedGraphNode *, UndirectedGraphNode *>::iterator it = f.find(node);
        if(it != f.end())   //if node is visited, just return the recorded nodeClone
            return f[node];

        UndirectedGraphNode *nodeClone = new UndirectedGraphNode(node->label);
        f[node] = nodeClone;
        for(vector<UndirectedGraphNode *>::size_type st = 0; st < node->neighbors.size(); st ++)
        {
            UndirectedGraphNode *temp = cloneGraph(node->neighbors[st]);
            if(temp != NULL)
                nodeClone->neighbors.push_back(temp);
        }
        return nodeClone;
    }
};