動機

複習dfs

Problem

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 value (int) and a list (List[Node]) of its neighbors.

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

 

Test case format:

For simplicity, 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.

An 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.

 

Example 1:

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).

Example 2:

Input: adjList = [[]]Output: [[]]Explanation: Note that the input contains one empty list. The graph consists of only one node with val = 1 and it does not have any neighbors.

Example 3:

Input: adjList = []Output: []Explanation: This an empty graph, it does not have any nodes.

Example 4:

Input: adjList = [[2],[1]]Output: [[2],[1]]

 

Constraints:

  • The number of nodes in the graph is in the range [0, 100].
  • 1 <= Node.val <= 100
  • Node.val is unique for each node.
  • There are no repeated edges and no self-loops in the graph.
  • The Graph is connected and all nodes can be visited starting from the given node.

Sol

要記得把生過的node拿出來用

class Solution:
    def __init__(self):
        self.mem = {}
    def cloneGraph(self, node: 'Node') -> 'Node':
        if not node:
            return None
        else:
            ret = Node(node.val)
            self.mem[node] = ret
            for n in node.neighbors:
                if n not in self.mem:
                    ret.neighbors.append(self.cloneGraph(n))
                else:
                    ret.neighbors.append(self.mem[n])
            return ret