leetcode || 133、Clone Graph

problem：

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

#

.

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
/ \
\_/

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题意：复制图（结构和数据不变，要新建节点）

thinking：

（1）要新建图的各个节点，维持邻接关系不变。

（2）采用unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> 存储原节点和新节点。而不是unordered_map<int, UndirectedGraphNode*>，

效率要高很多

（3）采用BFS思想，将原节点的邻接节点全部入栈或堆栈，遍历节点。

（4）map中查找key是否存在可以调用find（),也可以调用count（），后者效率更高

（5）提交没通过，结果不正确：

Input:{0,1,5#1,2,5#2,3#3,4,4#4,5,5#5}

Output:{0,5,1#1,5,2#2,3#3,4,4#4,5,5#5}

Expected:{0,1,5#1,2,5#2,3#3,4,4#4,5,5#5}

其实，结果是正确的，因为对于无向图，节点出现的顺序不影响图的结构，只能说这个验证程序只验证了一种结果

code：

class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> record;
if(node == NULL)
return node;

stack<UndirectedGraphNode*> queue;
queue.push(node);

while(!queue.empty()) {
UndirectedGraphNode *nextNode = queue.top();
queue.pop();

if(!record.count(nextNode)) {
UndirectedGraphNode *newNode = new UndirectedGraphNode(nextNode->label);
record[nextNode] = newNode;
}
for(int i = nextNode->neighbors.size()-1; i >= 0 ; i --) {
UndirectedGraphNode *childNode = nextNode->neighbors[i];
if(!record.count(childNode)) {
UndirectedGraphNode *newNode = new UndirectedGraphNode(childNode->label);
record[childNode] = newNode;
queue.push(childNode);
}
record[nextNode]->neighbors.push_back(record[childNode]);
}
}
return record[node];
}
};