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Clone Graph
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Change-Id: If70cd55492a20f1da193454f580ddfdb3d41126a
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applewjg committed Jan 14, 2015
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/*
Author: Andy, [email protected]
Date: Jan 13, 2015
Problem: Clone Graph
Difficulty: Medium
Source: http://oj.leetcode.com/problems/clone-graph/
Notes:
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 from 0 to N - 1, where N is the total nodes in the graph.
We use # as a separator for each node, and , as a separator for each neighbor of the node.
As an example, consider the serialized graph {1,2#2#2}.
The graph has a total of three nodes, and therefore contains three parts as separated by #.
Connect node 0 to both nodes 1 and 2.
Connect node 1 to node 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. DFS. 2. BFS.
*/

/**
* Definition for undirected graph.
* class UndirectedGraphNode {
* int label;
* List<UndirectedGraphNode> neighbors;
* UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
* };
*/
public class Solution {
public UndirectedGraphNode cloneGraph_1(UndirectedGraphNode node) {
HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
return cloneGraphRe(node, map);
}
public UndirectedGraphNode cloneGraphRe(UndirectedGraphNode node, HashMap<UndirectedGraphNode, UndirectedGraphNode> map) {
if (node == null) return null;
if (map.containsKey(node) == true) {
return map.get(node);
}
UndirectedGraphNode newnode = new UndirectedGraphNode(node.label);
map.put(node, newnode);
for (UndirectedGraphNode cur : node.neighbors) {
newnode.neighbors.add(cloneGraphRe(cur, map));
}
return newnode;
}
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
if (node == null) return null;
queue.offer(node);
map.put(node, new UndirectedGraphNode(node.label));
while (queue.isEmpty() == false) {
UndirectedGraphNode cur = queue.poll();
for (UndirectedGraphNode neighbor : cur.neighbors) {
if (map.containsKey(neighbor) == false) {
UndirectedGraphNode newnode = new UndirectedGraphNode(neighbor.label);
map.put(neighbor, newnode);
queue.offer(neighbor);
}
map.get(cur).neighbors.add(map.get(neighbor));
}
}
return map.get(node);
}
}

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