leetcode Clone Graph

复制一个无向图。图的结构时有一个label，一个vector存和他想接的节点。可以自循环，就是vector中可以存在自己。例如：

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：一开始，我就想到用BFS，但是需要判断是否已经处理过该节点，因为存在循环。避免死循环就要用一个可以O1时间访问否值是否存在的东东来存已经访问过的节点，那就是map了，或者是set，我用set存访问过的，然后两个queue来同步的走，及把node复制到另一个root中，然后que和que2分别表示需要继续BFS处理的后序节点。非迭代的：

/**
 * 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 node;
        //if ((node -> neighbors).size() == 0) return node;
        queue<UndirectedGraphNode *> que, que2;
        que.push(node);
        UndirectedGraphNode *root = new UndirectedGraphNode(node -> label), *tmp, *subnode, *tmp2;
        unordered_set<UndirectedGraphNode*> uset;
        que2.push(root);
        while(!que.empty())
        {
            tmp = que.front();
            tmp2 = que2.front();
            uset.insert(tmp);
            que.pop();
            que2.pop();
            for (int i = 0; i < tmp -> neighbors.size(); ++i)
            {
                if (tmp -> neighbors[i] == tmp)
                    subnode = tmp2;
                else
                    subnode = new UndirectedGraphNode(tmp -> neighbors[i] -> label);
                tmp2 -> neighbors.push_back(subnode);
                if (uset.count(tmp -> neighbors[i]) == 0)
                {
                    que.push(tmp -> neighbors[i]);
                    que2.push(subnode);
                }
            }
        }
        return root;
    }
};

但是超时了，可能最近感冒脑子不灵活了哎。

不知道为什么上面用两个队列会超时啊。换成map的映射的话就不会超时：

unordered_map(UndirectedGraphNode *, UndirectedGraphNode *) copied; 指key对应的node是否在value对应的node中复制。

因为是BFS要用一个que来存自己已经复制，但是他的neighbors还没有复制的节点。那么初始化que就要push一个node，而copied[node]就是赋值为node->label对应的新节点。

然后根据que是否为空处理。

/**
 * 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 node;
        unordered_map<UndirectedGraphNode *, UndirectedGraphNode *> copied;
        queue<UndirectedGraphNode *> que;
        que.push(node);
        copied[node] = new UndirectedGraphNode(node -> label);
        
        while(!que.empty())
        {
            UndirectedGraphNode *cur = que.front();
            que.pop();
            for (int i = 0; i < cur -> neighbors.size(); ++i)
            {
                if (copied.count(cur->neighbors[i]))
                    copied[cur] -> neighbors.push_back(copied[cur -> neighbors[i]]);
                else
                {
                    UndirectedGraphNode *new_node = new UndirectedGraphNode(cur -> neighbors[i] -> label);
                    copied[cur -> neighbors[i]] = new_node;
                    copied[cur] -> neighbors.push_back(new_node);
                    que.push(cur -> neighbors[i]);
                }
            }
        }
        return copied[node];
    }
};

 这有DFS