Leetcode 133. Clone Graph

Description: 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.

 Link: 133. Clone Graph

Examples:

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]]

思路: deep copy，就是要所有node重新创建，关系重新连接好。

class Solution(object):
    def cloneGraph(self, node):
        """
        :type node: Node
        :rtype: Node
        """
        return self.dfs(node, dict())
       
    
    def dfs(self, node, nodes):
        if not node: return None
        if node.val in nodes: return nodes[node.val]
        new = Node(node.val, [])
        nodes[node.val] = new
        for n in node.neighbors:
            new_n = self.dfs(n, nodes)
            if new_n:
                new.neighbors.append(new_n)
        return new

日期: 2021-03-26