[Leetcode]Clone Graph

最新推荐文章于 2019-07-20 20:47:16 发布

NoooName

最新推荐文章于 2019-07-20 20:47:16 发布

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分类专栏： * graph 文章标签： leetcode python

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本文介绍了一种用于复制无向图的方法，利用深度优先搜索（DFS）和广度优先搜索（BFS）。通过这两种算法可以有效地创建原始图的一个完全副本，包括所有节点及其连接。

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Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

OJ's undirected graph serialization:

Nodes are labeled uniquely.

We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.

As an example, consider the serialized graph {0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by #.

First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
Second node is labeled as 1. Connect node 1 to node 2.
Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

Visually, the graph looks like the following:

图的复制~可以用深度优先和广度优先来做~深度优先代码如下~

class Solution:
    # @param node, a undirected graph node
    # @return a undirected graph node
    def cloneGraph(self, node):
        if node is None: return None
        dict = {}
        return self.dfs(node, dict)
        
    def dfs(self, node, dict):
        if node in dict:
            return dict[node]
        nodeCopy = UndirectedGraphNode(node.label)
        dict[node] = nodeCopy
        for neighbor in node.neighbors:
            nodeCopy.neighbors.append(self.dfs(neighbor, dict))
        return nodeCopy

广度优先代码如下

class Solution:
    # @param node, a undirected graph node
    # @return a undirected graph node
    def cloneGraph(self, node):
        if node is None: return None
        dict = {}
        queue = collections.deque()
        queue.append(node)
        nodeCopy = UndirectedGraphNode(node.label)
        dict[node] = nodeCopy
        while queue:
            curr = queue.popleft()
            for neighbor in curr.neighbors:
                if neighbor in dict:
                    dict[curr].neighbors.append(dict[neighbor])
                else:
                    neighborCopy = UndirectedGraphNode(neighbor.label)
                    dict[curr].neighbors.append(neighborCopy)
                    dict[neighbor] = neighborCopy
                    queue.append(neighbor)
        return nodeCopy