• 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.

    Example:

    Input:
    {"$id":"1","neighbors":[{"$id":"2","neighbors":[{"$ref":"1"},{"$id":"3","neighbors":[{"$ref":"2"},{"$id":"4","neighbors":[{"$ref":"3"},{"$ref":"1"}],"val":4}],"val":3}],"val":2},{"$ref":"4"}],"val":1}
    
    Explanation:
    Node 1's value is 1, and it has two neighbors: Node 2 and 4.
    Node 2's value is 2, and it has two neighbors: Node 1 and 3.
    Node 3's value is 3, and it has two neighbors: Node 2 and 4.
    Node 4's value is 4, and it has two neighbors: Node 1 and 3.

    Note:

    1. The number of nodes will be between 1 and 100.
    2. The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph.
    3. Since the graph is undirected, if node p has node q as neighbor, then node q must have node p as neighbor too.
    4. You must return the copy of the given node as a reference to the cloned graph.

    Approach #1: DFS. [Java]

    /*
    // Definition for a Node.
    class Node {
        public int val;
        public List<Node> neighbors;
    
        public Node() {}
    
        public Node(int _val,List<Node> _neighbors) {
            val = _val;
            neighbors = _neighbors;
        }
    };
    */
    class Solution {
        public Node clone(Node node, HashMap<Integer, Node> map) {
            if (node == null) return null;
            if (map.containsKey(node.val))
                return map.get(node.val);
            Node newNode = new Node(node.val, new ArrayList<Node>());
            map.put(newNode.val, newNode);
            for (Node neighbor : node.neighbors)
                newNode.neighbors.add(clone(neighbor, map));
            return newNode;
        }
        
        public Node cloneGraph(Node node) {
            HashMap<Integer, Node> map = new HashMap<>();
            return clone(node, map);
        }
    }
    

      

    Analysis:

    https://leetcode.com/problems/clone-graph/discuss/42309/Depth-First-Simple-Java-Solution

    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/10952288.html
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