Graph 133. Clone Graph in three ways(bfs, dfs, bfs(recursive))
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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 / \\_/
Basically just clone the graph like clone a list in leetcode 138.
there are three ways t solve this (just traverse the graph and put new node into map)
/** * Definition for undirected graph. * class UndirectedGraphNode { * int label; * List<UndirectedGraphNode> neighbors; * UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); } * }; */ public class Solution { //dfs Map<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>(); public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) { if(node==null) return null; //copy graph(deep copy), hashmap map.put(node, new UndirectedGraphNode(node.label)); helper(node); return map.get(node); } void helper(UndirectedGraphNode node){ for(int i = 0; i< node.neighbors.size(); i++){ UndirectedGraphNode neighbor = node.neighbors.get(i); if(!map.containsKey(neighbor)){// not visited UndirectedGraphNode newNode = new UndirectedGraphNode(neighbor.label); map.put(neighbor, newNode);//visited helper(neighbor);//why put helper here } map.get(node).neighbors.add(map.get(neighbor)); //set the link of neighbors } } }
Solution 2: bfs queue
/** * 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(UndirectedGraphNode node) { if(node==null) return null; //bfs LinkedList<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>(); queue.offer(node); Map<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>(); UndirectedGraphNode newNode = new UndirectedGraphNode(node.label); map.put(node,newNode); while(!queue.isEmpty()){ UndirectedGraphNode cur = queue.poll();//pop for(int i = 0; i<cur.neighbors.size(); i++){ UndirectedGraphNode neighbor = cur.neighbors.get(i); if(!map.containsKey(neighbor)){ queue.offer(neighbor); newNode = new UndirectedGraphNode(neighbor.label); map.put(neighbor, newNode); map.get(cur).neighbors.add(newNode); } //if contains the key else map.get(cur).neighbors.add(map.get(neighbor)); } } return map.get(node); } }
Solution 3: dfs with all node connected.
/** * 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(UndirectedGraphNode node) { if(node == null) return null; //dfs, if not visited, visited it and set it to visited, stack LinkedList<UndirectedGraphNode> stack = new LinkedList<>();//add first stack.push(node); Map<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<>(); UndirectedGraphNode newNode = new UndirectedGraphNode(node.label); map.put(node, newNode); while(!stack.isEmpty()){ UndirectedGraphNode cur = stack.pop();//pop for(int i = 0; i<cur.neighbors.size(); i++){ UndirectedGraphNode neighbor = cur.neighbors.get(i);//neighbor of current if(!map.containsKey(neighbor)){//put neighbor into hashmap (visited) newNode = new UndirectedGraphNode(neighbor.label);//copy neighbors map.put(neighbor, newNode); stack.push(neighbor); } //set the link of neighbors map.get(cur).neighbors.add(map.get(neighbor)); } } return map.get(node); } } // relationship in hashmap // key, value // cur, map.get(cur) // cur.neighbors, newNode/ map.get(eighbor)
What if nodes are not connnected partly: just write a loop to chekc all the node(call dfs for each node) in the graph
https://www.geeksforgeeks.org/depth-first-search-or-dfs-for-a-graph/
How do you represent the graph(one way from leetcode, another from geekforgeek)
Lastly: think about the time complexity of them
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