LeetCode每日一题(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 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.

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.

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早期做这个题，提交了十几次，不是结果不对就是超时，现在回来看这个题，觉得经验真的很重要，当时一门心思的走 DFS 的路子，倒也不是不能做出来，但是代码会被搞的极为复杂，搞到最后自己可能猜到问题在哪了，但是改了之后会出现另一个问题。现在回过头来看，可能大部分的设计 Graph 的问题，BFS 才是更好而且更符合直觉的解决方式。这题用了两次 BFS 的遍历，一次复制 Node 本身， 一次给这些复制的 Node 添加关联。

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func cloneGraph(node *Node) *Node {
	if node == nil {
		return nil
	}
	nexts := make([]*Node, 0, 100)
	nexts = append(nexts, node)
	nodes := make(map[int]*Node)
	visited := make(map[int]bool)
	for len(nexts) > 0 {
		n, remain := nexts[0], nexts[1:]
		nexts = remain
		if visited[n.Val] {
			continue
		}
		visited[n.Val] = true
		nodes[n.Val] = &Node{Val: n.Val}
		for _, neighbor := range n.Neighbors {
			if visited[n.Val] {
				nexts = append(nexts, neighbor)
			}
		}
	}
	visited = make(map[int]bool)
	nexts = make([]*Node, 0, 100)
	nexts = append(nexts, node)
	for len(nexts) > 0 {
		n, remain := nexts[0], nexts[1:]
		nexts = remain
		if visited[n.Val] {
			continue
		}
		visited[n.Val] = true
		for _, neighbor := range n.Neighbors {
			nodes[n.Val].Neighbors = append(nodes[n.Val].Neighbors, nodes[neighbor.Val])
		}
		for _, neighbor := range n.Neighbors {
			if visited[n.Val] {
				nexts = append(nexts, neighbor)
			}
		}
	}
	return nodes[node.Val]
}