试题
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.
Example:
Input:
{“KaTeX parse error: Expected '}', got 'EOF' at end of input: …"neighbors":[{"id”:“2”,“neighbors”:[{“KaTeX parse error: Expected 'EOF', got '}' at position 9: ref":"1"}̲,{"id”:“3”,“neighbors”:[{“KaTeX parse error: Expected 'EOF', got '}' at position 9: ref":"2"}̲,{"id”:“4”,“neighbors”:[{“KaTeX parse error: Expected 'EOF', got '}' at position 9: ref":"3"}̲,{"ref”:“1”}],“val”:4}],“val”:3}],“val”:2},{"$ref":“4”}],“val”:1}
Explanation:
Node 1’s value is 1, and it has two neighbors: Node 2 and 4.
Node 2’s value is 2, and it has two neighbors: Node 1 and 3.
Node 3’s value is 3, and it has two neighbors: Node 2 and 4.
Node 4’s value is 4, and it has two neighbors: Node 1 and 3.
代码
克隆时候要保证相同节点只是被新建了一次,幸好val-1的值可以作为节点标号。
/*
// Definition for a Node.
class Node {
public int val;
public List<Node> neighbors;
public Node() {}
public Node(int _val,List<Node> _neighbors) {
val = _val;
neighbors = _neighbors;
}
};
*/
class Solution {
public Node cloneGraph(Node node) {
if(node == null) return node;
Node copyNode = new Node(node.val, new ArrayList<Node>());
Node[] nodes = new Node[100];
nodes[node.val-1] = copyNode;
dfs(copyNode, node, nodes);
return copyNode;
}
public void dfs(Node copyNode, Node node, Node[] nodes) {
for(Node n : node.neighbors){
if(nodes[n.val-1] != null){
copyNode.neighbors.add(nodes[n.val-1]);
continue;
}
Node temp = new Node(n.val, new ArrayList<Node>());
nodes[n.val-1] = temp;
copyNode.neighbors.add(temp);
dfs(temp, n, nodes);
}
}
}