leetcode解题: Clone Graph (133)

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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:

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   1
  / \
 /   \
0 --- 2
     / \
     \_/

解法1：HashMap + DFS, O(N) Time + O(N) Space

思路和有一题linkedlist有random pointer的一样，要分两步进行，一个是要遍历整个图（可选dfs或者bfs），另外要用一个hashmap来记录node和node之间的对应关系。

Java

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/**
 * 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 node;
        }
        
        HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
        dfs(node, map);
        for (UndirectedGraphNode old: map.keySet()) {
            if (node.neighbors != null) {
                List<UndirectedGraphNode> ns = new ArrayList<UndirectedGraphNode>();
                for (int i = 0; i < old.neighbors.size(); ++i) {
                    ns.add(map.get(old.neighbors.get(i)));
                }
                map.get(old).neighbors = ns;
            }
        }
        
        return map.get(node);
    }
    
    void dfs(UndirectedGraphNode node, HashMap<UndirectedGraphNode, UndirectedGraphNode> map) {
        if (node == null) {
            return;
        }
        if (!map.containsKey(node)) {
            // Delay the assignment of neighbors
            UndirectedGraphNode cloned = new UndirectedGraphNode(node.label);
            map.put(node, cloned);
            if (node.neighbors!= null) {
                for (int i = 0; i < node.neighbors.size(); ++i) {
                    dfs(node.neighbors.get(i), map);
                }
            }
        }
    }
}