[leetcode] 133. Clone Graph 解题报告

题目链接：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 #.

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
         / \
         \_/

思路：克隆图的一个难点就是一个结点的邻居可能在已经出现过，这样你只要把他的指针加到邻居集合中即可，也有可能这个结点还没出现过，因此你需要新建一个这个结点，因此我们需要一个hash表来对结点做一一映射．

本题有两种方法来做，广度搜索BFS和深度搜索DFS.

1. 广度搜索BFS

利用队列，每次遍历队列头结点所有的邻居，如果其某个邻居还没有被创建，那就创建一个，并且将其加入到队列中去．否则说明这个结点已经被创建，而我们只在一个结点刚被创建时才将其加入到队列中去，因此这个结点已经被创建我们就没必要将其入队列了．每次遍历完之后就克隆好了一个结点和其邻居．然后直到队列为空，我们就完成了克隆的操作．

代码如下：

/**
 * 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;
        UndirectedGraphNode * graph = new UndirectedGraphNode(node->label);
        hash[node] = graph;
        queue<UndirectedGraphNode*> que;
        que.push(node);
        while(!que.empty())
        {
            auto curNode = que.front();
            que.pop();
            for(auto val: curNode->neighbors)
            {
                if(hash.find(val) == hash.end())
                {
                    hash[val] = new UndirectedGraphNode(val->label);
                    que.push(val);
                }
                hash[curNode]->neighbors.push_back(hash[val]);
            }
        }
        return graph;
    }
private:
    unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> hash;
};

２．深度搜索DFS

DFS利用递归一般可以写出比BFS更优雅整洁的代码．

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．

代码如下：

/**
 * 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.find(node) == hash.end())
        {
            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;
};

参考：https://leetcode.com/discuss/41944/7-17-lines-c-bfs-dfs-solutions