LeetCode: Clone Graph 解题报告
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 #.
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
/ \
\_/
Solution 1:
使用BFS来解决此问题。用一个Queue来记录遍历的节点,遍历原图,并且把复制过的节点与原节点放在MAP中防止重复访问。
图的遍历有两种方式,BFS和DFS
这里使用BFS来解本题,BFS需要使用queue来保存neighbors
但这里有个问题,在clone一个节点时我们需要clone它的neighbors,而邻居节点有的已经存在,有的未存在,如何进行区分?
这里我们使用Map来进行区分,Map的key值为原来的node,value为新clone的node,当发现一个node未在map中时说明这个node还未被clone,
将它clone后放入queue中处理neighbors。
使用Map的主要意义在于充当BFS中Visited数组,它也可以去环问题,例如A--B有条边,当处理完A的邻居node,然后处理B节点邻居node时发现A已经处理过了
处理就结束,不会出现死循环。
queue中放置的节点都是未处理neighbors的节点。
http://www.cnblogs.com/feiling/p/3351921.html
/*
Iteration Solution:
*/
public UndirectedGraphNode cloneGraph1(UndirectedGraphNode node) {
if (node == null) {
return null;
} UndirectedGraphNode root = null; // store the nodes which are cloned.
HashMap<UndirectedGraphNode, UndirectedGraphNode> map =
new HashMap<UndirectedGraphNode, UndirectedGraphNode>(); Queue<UndirectedGraphNode> q = new LinkedList<UndirectedGraphNode>(); q.offer(node);
UndirectedGraphNode rootCopy = new UndirectedGraphNode(node.label); // 别忘记这一行啊。orz..
map.put(node, rootCopy); // BFS the graph.
while (!q.isEmpty()) {
UndirectedGraphNode cur = q.poll();
UndirectedGraphNode curCopy = map.get(cur); // bfs all the childern node.
for (UndirectedGraphNode child: cur.neighbors) {
// the node has already been copied. Just connect it and don't need to copy.
if (map.containsKey(child)) {
curCopy.neighbors.add(map.get(child));
continue;
} // put all the children into the queue.
q.offer(child); // create a new child and add it to the parent.
UndirectedGraphNode childCopy = new UndirectedGraphNode(child.label);
curCopy.neighbors.add(childCopy); // Link the new node to the old map.
map.put(child, childCopy);
}
} return rootCopy;
}
2014.12.30 Redo:
/*
SOLUTION 3: The improved Version.
*/
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
if (node == null) {
return null;
} HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>(); // BUG 1: can't use queue , should use LinkedList.
Queue<UndirectedGraphNode> q = new LinkedList<UndirectedGraphNode>(); q.offer(node); // copy the root node. and then put it into the map.
UndirectedGraphNode nodeCopy = new UndirectedGraphNode(node.label);
map.put(node, nodeCopy); while (!q.isEmpty()) {
UndirectedGraphNode cur = q.poll(); // get out the copy node. We guarantee that it has been copied. Because we always put it into the map before
// put it into the queue.
UndirectedGraphNode curCopy = map.get(cur); // go through all the children node.
// Line 71: java.util.ConcurrentModificationException. use cur instead of curCopy
for (UndirectedGraphNode child: cur.neighbors) { if (map.containsKey(child)) {
curCopy.neighbors.add(map.get(child));
} else {
// Only add the child into the map when it is not visited.
q.offer(child); // BUG 3: forget to add the new node into the map.
UndirectedGraphNode childCopy = new UndirectedGraphNode(child.label);
curCopy.neighbors.add(childCopy);
map.put(child, childCopy);
}
}
} return map.get(node);
}
Solution 2:
同样的,我们也可以使用递归DFS来解决此题,思路与上图一致,但为了避免重复运算产生死循环。当进入DFS时,如果发现map中已经有了拷贝过的值,直接退出即可。
题目虽然简单,但主页君仍然考虑了递归的特性使程序简洁。比如:我们拷贝只拷贝根节点,而子节点的拷贝由recursion来完成,这样可以使程序更加简洁。
注意:要先加入到map,再调用rec ,否则会造成不断地反复拷贝而死循环。
/*
Solution 2: Recursion version.
*/
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
if (node == null) {
return null;
} return rec(node, new HashMap<UndirectedGraphNode, UndirectedGraphNode>());
} public UndirectedGraphNode rec(UndirectedGraphNode root, HashMap<UndirectedGraphNode, UndirectedGraphNode> map) {
// If it has been copied, just return the copy node from the map.
UndirectedGraphNode rootCopy = map.get(root);
if (rootCopy != null) {
return rootCopy;
} // if the root is not copied, create a new one.
rootCopy = new UndirectedGraphNode(root.label);
map.put(root, rootCopy); // copy all the child node.
for (UndirectedGraphNode child: root.neighbors) {
// call the recursion to create all the children and add the new children to the copy node.
rootCopy.neighbors.add(rec(child, map));
} return rootCopy;
}
2014.12.30 Redo:
public UndirectedGraphNode cloneGraph1(UndirectedGraphNode node) {
if (node == null) {
return null;
} return rec(node, new HashMap<UndirectedGraphNode, UndirectedGraphNode>());
} // SOLUTION 1:
// Try to return a copied cloneGraph.
public UndirectedGraphNode rec(UndirectedGraphNode node, HashMap<UndirectedGraphNode, UndirectedGraphNode> map) {
// The base case:
if (map.containsKey(node)) {
// If the map has been copied, just return the node.
return map.get(node);
} // create a new node.
UndirectedGraphNode nodeCopy = new UndirectedGraphNode(node.label);
// BUG 2: should put it into the map first. Because we don't want to copy the same node again in the recursion. map.put(node, nodeCopy);
for (int i = 0; i < node.neighbors.size(); i++) {
// BUG 1: forget a parameter.
// copy all the children node.
nodeCopy.neighbors.add(rec(node.neighbors.get(i), map));
} return nodeCopy;
}
Ref: http://m.blog.csdn.net/blog/hellobinfeng/17497883
Code:
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