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

#

.

First node is labeled as

0

. Connect node

0

to both nodes

1

and

2

.

Second node is labeled as

1

. Connect node

1

to node

2

.

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;
}
};

View Code
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