133. 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<Node> 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).
My Solutions:
class Solution {
public Node cloneGraph(Node node) {
if (node == null) return null;
Node copy = new Node(node.val);
Node[] visited = new Node[101];
dfs(node, copy, visited);
return copy;
}
private void dfs(Node node, Node copy, Node[] visited) {
// store the current node at it's val index which will tell us that this node is now visited
visited[copy.val] = copy;
// 遍历每个neighbors
for (Node n : node.neighbors) {
// this node has not been visited
if (visited[n.val] == null) {
Node newNode = new Node(n.val);
copy.neighbors.add(newNode);
dfs(n, newNode, visited);
}
// if this node has been visited, retrieve node from visited array
// and add it as the adjacent node of the copied node
else {
// THIS IS THE POINT WHY WE USED NODE[] INSTEAD OF BOOLEAN[] ARRAY
copy.neighbors.add(visited[n.val]);
}
}
}
}
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