[LeetCode] 133. Clone Graph

Given a reference of a node in a connected undirected graph.

Return a deep copy (clone) of the graph.

Each node in the graph contains a val (int) and a list (List[Node]) of its neighbors.

class Node {
    public int val;
    public List<Node> neighbors;
}

Test case format:

For simplicity sake, each node's value is the same as the node's index (1-indexed). For example, the first node with val = 1, the second node with val = 2, and so on. The graph is represented in the test case using an adjacency list.

Adjacency list is a collection of unordered lists used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.

The given node will always be the first node with val = 1. You must return the copy of the given node as a reference to the cloned graph.

class Node {
    public int val;
    public List<Node> neighbors;
}

Input: adjList = [[2,4],[1,3],[2,4],[1,3]]
Output: [[2,4],[1,3],[2,4],[1,3]]
Explanation: There are 4 nodes in the graph.
1st node (val = 1)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
2nd node (val = 2)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).
3rd node (val = 3)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
4th node (val = 4)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).

克隆图。

题目就是题意，对一个图做深度复制。这里需要被复制的内容不光是图上的每个节点，同时也需要为每个节点的邻接点做深度复制。

既然是克隆/复制一个数据结构，一定涉及到两点，一是遍历，二是记录。图的遍历无非就是BFS或者DFS，记录基本都会用到hashmap，除了链表的复制也许可以不需要用到hashmap，因为可以把复制出来的节点加在原节点之后。参见138题。

这个题也可以用DFS做但是我个人觉得BFS比较好记。具体的做法是创建一个visited hashmap记住<原来的node，新的node>之间的对应关系，和一个queue。

一开始，先把当前唯一的节点node加入queue并且把当前节点当做key存入hashmap，map的value是一个有着相同node.val但是只有一个空的邻接表（neighbors）的node；

然后从queue中开始弹出元素，每弹出一个元素cur，判断这个node的每一个邻居是否存在于hashmap，将不存在的邻居加入hashmap，同时也将这些刚刚加入hashmap中的邻居再次加入queue进行下一轮遍历；如果有某一个邻居已经存在于hashmap了，则往cur的邻接表里面添加这个已经存在的邻居的copy。

BFS

时间O(V + E)

空间O(n) - queue

Java实现

/*
class Node {
    public int val;
    public List<Node> neighbors;
    
    public Node() {
        val = 0;
        neighbors = new ArrayList<Node>();
    }
    
    public Node(int _val) {
        val = _val;
        neighbors = new ArrayList<Node>();
    }
    
    public Node(int _val, ArrayList<Node> _neighbors) {
        val = _val;
        neighbors = _neighbors;
    }
}
*/

class Solution {
    public Node cloneGraph(Node node) {
        // corner case
        if (node == null) {
            return null;
        }
        // normal case
        HashMap<Node, Node> visited = new HashMap<>();
        Queue<Node> queue = new LinkedList<>();
        queue.add(node);
        // <oldNode, newNode>
        visited.put(node, new Node(node.val, new ArrayList<>()));
        while (!queue.isEmpty()) {
            Node cur = queue.poll();
            for (Node neighbor : cur.neighbors) {
                if (!visited.containsKey(neighbor)) {
                    visited.put(neighbor, new Node(neighbor.val, new ArrayList()));
                    queue.offer(neighbor);
                }
                // add a copy of this neighbor to the copy of cur
                visited.get(cur).neighbors.add(visited.get(neighbor));
            }
        }
        return visited.get(node);
    }
}

DFS

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public Node cloneGraph(Node node) {
 3         HashMap<Node, Node> map = new HashMap<>();
 4         return dfs(node, map);
 5     }
 6 
 7     private Node dfs(Node node, HashMap<Node, Node> map) {
 8         if (node == null) {
 9             return null;
10         }
11         if (map.containsKey(node)) {
12             return map.get(node);
13         }
14         Node clone = new Node(node.val, new ArrayList<>());
15         map.put(node, clone);
16         for (Node neighbor : node.neighbors) {
17             clone.neighbors.add(dfs(neighbor, map));
18         }
19         return clone;
20     }
21 }