[LintCode] 137 Clone Graph 解题报告

Description
Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

How we serialize an undirected graph:

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
       / \
       \_/



Example
return a deep copied graph.



思路
首先会想到用BFS或者DFS去遍历这个图，然后对于每一个点，我们做一个新的点出来。我们要解决的问题是新做的点之间要像遍历过的点那样连起来。由此我们想到用HashMap来做老的点和新的点之间的对应。
我们用BFS做。具体看代码。



Code

/**
 * Definition for undirected graph.
 * class UndirectedGraphNode {
 *     int label;
 *     ArrayList<UndirectedGraphNode> neighbors;
 *     UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
 * };
 */
public class Solution {
    /**
     * @param node: A undirected graph node
     * @return: A undirected graph node
     */
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        // write your code here
        
        if (node == null) {
            return null;
        }
        
        HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<>();
        Queue<UndirectedGraphNode> queue = new LinkedList<>();
        
        queue.offer(node);
        map.put(node, new UndirectedGraphNode(node.label));
        
        while (!queue.isEmpty()) {
            UndirectedGraphNode currNode = queue.poll();
            for (UndirectedGraphNode neighbor : currNode.neighbors) {
                if (!map.containsKey(neighbor)) {
                    map.put(neighbor, new UndirectedGraphNode(neighbor.label));
                    queue.offer(neighbor);
                }
                map.get(currNode).neighbors.add(map.get(neighbor));
            }
        }
        
        return map.get(node);
    }
}