Clone Graph Leetcode

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 #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. 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
         / \
         \_/

 

这道题可以用BFS.

要注意:

clone的时候，指向的点也要是新建的点，不可指向原来的点;

返回的点要是原来的node所复制出来的点。

/**
 * 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;
        }
        Map<Integer, UndirectedGraphNode> hm = new HashMap<>();
        Queue<UndirectedGraphNode> q = new LinkedList<>();
        q.add(node);
        
        UndirectedGraphNode clone = new UndirectedGraphNode(node.label);
        hm.put(node.label, clone);
        UndirectedGraphNode tmp = null;
        while (!q.isEmpty()) {
            UndirectedGraphNode no = q.poll();
            tmp = hm.getOrDefault(no.label, new UndirectedGraphNode(no.label));
            for (UndirectedGraphNode n : no.neighbors) {
                UndirectedGraphNode cloneN = null;
                if (hm.containsKey(n.label)) {
                    cloneN = hm.get(n.label);
                } else {
                    cloneN = new UndirectedGraphNode(n.label);
                    q.add(n);
                    hm.put(n.label, cloneN);
                }
                tmp.neighbors.add(cloneN);
            }
        }
        return clone;
    }
}

开心的是和top solution思路一样，还是按照他们的改了一下，虽然我觉得我的思路最清晰了。。。

期间发现不用getOrDefault因为一定会有。

public class Solution {
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if (node == null) {
            return null;
        }
        Map<Integer, UndirectedGraphNode> hm = new HashMap<>();
        Queue<UndirectedGraphNode> q = new LinkedList<>();
        q.add(node);
        
        UndirectedGraphNode clone = new UndirectedGraphNode(node.label);
        hm.put(node.label, clone);
        while (!q.isEmpty()) {
            UndirectedGraphNode no = q.poll();
            for (UndirectedGraphNode n : no.neighbors) {
                if (!hm.containsKey(n.label)) {
                    q.add(n);
                    hm.put(n.label, new UndirectedGraphNode(n.label));
                }
                hm.get(no.label).neighbors.add(hm.get(n.label));
            }
        }
        return clone;
    }
}

 

DFS以后再写吧。。。。到时候再回顾。。。