标签：leetcode 图 dfs bfs unordered_map
problem：
Clone an undirected graph. Each node in the graph contains a label and
a list of its neighbors.
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 #.
0.
Connect node 0 to both nodes 1 and 2.1.
Connect node 1 to node 2.2.
Connect node 2 to node 2 (itself),
thus forming a self-cycle.Visually, the graph looks like the following:
1
/ / 0 --- 2
/ \_/
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];
}
};标签：leetcode 图 dfs bfs unordered_map
原文地址：http://blog.csdn.net/hustyangju/article/details/45363113
