[LeetCode] 133. FIG clone

Topic links: https://leetcode-cn.com/problems/clone-graph/

Subject description:

Given an undirected communication reference drawing of a node, return the figure deep copy (clone). Each node in the graph contains its value val( Int), and a list of its neighbors ( list[Node]).

Example:

输入：
{"$id":"1","neighbors":[{"$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}

解释：
节点 1 的值是 1，它有两个邻居：节点 2 和 4 。
节点 2 的值是 2，它有两个邻居：节点 1 和 3 。
节点 3 的值是 3，它有两个邻居：节点 2 和 4 。
节点 4 的值是 4，它有两个邻居：节点 1 和 3 。

prompt:

• Nodes between 1 and 100.
• Undirected graph is a simple graph, which means that there are no duplicate FIG side, since there is no loop.
• As the graph is undirected, if the node is p q neighbor node, the node node p q must also be neighbors.
• Copied node must be given as a reference to returns clones FIG.

Ideas:

This question is to traverse the entire map, so the time to traverse the access point has been recorded, we use a dictionary record

So, there are two traversal methods

A thought: DFS (depth traversal)

Thinking two: BFS (breadth traversal)

Code:

A thought:

class Solution:
    def cloneGraph(self, node: 'Node') -> 'Node':
        lookup = {}

        def dfs(node):
            #print(node.val)
            if not node: return
            if node in lookup:
                return lookup[node]
            clone = Node(node.val, [])
            lookup[node] = clone
            for n in node.neighbors:
                clone.neighbors.append(dfs(n))
            
            return clone

        return dfs(node)

java

class Solution {
    public Node cloneGraph(Node node) {
        Map<Node, Node> lookup = new HashMap<>();
        return dfs(node, lookup);
    }

    private Node dfs(Node node, Map<Node,Node> lookup) {
        if (node == null) return null;
        if (lookup.containsKey(node)) return lookup.get(node);
        Node clone = new Node(node.val, new ArrayList<>());
        lookup.put(node, clone);
        for (Node n : node.neighbors)clone.neighbors.add(dfs(n,lookup));
        return clone;
    }
}

Thinking two:

"""
# Definition for a Node.
class Node:
    def __init__(self, val, neighbors):
        self.val = val
        self.neighbors = neighbors
"""
class Solution:
    def cloneGraph(self, node: 'Node') -> 'Node':
        from collections import deque
        lookup = {}

        def bfs(node):
            if not node: return
            clone = Node(node.val, [])
            lookup[node] = clone
            queue = deque()
            queue.appendleft(node)
            while queue:
                tmp = queue.pop()
                for n in tmp.neighbors:
                    if n not in lookup:
                        lookup[n] = Node(n.val, [])
                        queue.appendleft(n)
                    lookup[tmp].neighbors.append(lookup[n])
            return clone

        return bfs(node)

java

class Solution {
    public Node cloneGraph(Node node) {
        if (node == null) return null;
        Map<Node, Node> lookup = new HashMap<>();
        Node clone = new Node(node.val, new ArrayList<>());
        lookup.put(node, clone);
        Deque<Node> queue = new LinkedList<>();
        queue.offer(node);
        while (!queue.isEmpty()) {
            Node tmp = queue.poll();
            for (Node n : tmp.neighbors) {
                if (!lookup.containsKey(n)) {
                    lookup.put(n, new Node(n.val, new ArrayList<>()));
                    queue.offer(n);
                }
                lookup.get(tmp).neighbors.add(lookup.get(n));
            }
        }
        return clone;
    }
}

Source: stay button (LeetCode)
link: https://leetcode-cn.com/problems/clone-graph
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