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 #.

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
         / \
         \_/

广度优先搜索法

思路

用一个hashmap将原节点和复制的节点映射起来, 因为hashmap的存在 可以不用记录visited, 因为凡是便利过的都在map中

复杂度

一次便利时间O(n) 空间O(n)

代码

public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
    if (node == null) {
        return null;
    }
    HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
    map.put(node, new UndirectedGraphNode(node.label));
    Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
    queue.offer(node);
    while (!queue.isEmpty()) {
        UndirectedGraphNode cur = queue.poll();
        for (UndirectedGraphNode tem : cur.neighbors) {
            if (!map.containsKey(tem)) {
                queue.offer(tem);
                map.put(tem, new UndirectedGraphNode(tem.label));
            }
            map.get(cur).neighbors.add(map.get(tem));
        }
    }
    return map.get(node);
}

深度优先搜索

复杂度

时间O(n) 空间O(n)

代码

public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
    if (node == null) {
        return null;
    }
    HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
    map.put(node, new UndirectedGraphNode(node.label));
    dfs(map, node);
    return map.get(node);
}
public void dfs(HashMap<UndirectedGraphNode, UndirectedGraphNode> map, UndirectedGraphNode node) {
    if (node == null) {
        return;
    }
    for (UndirectedGraphNode cur : node.neighbors) {
        if (!map.containsKey(cur)) {
            map.put(cur, new UndirectedGraphNode(cur.label));
            dfs(map, cur);
        }
        map.get(node).neighbors.add(map.get(cur));
    }
}