leetcode 133) Clone Graph

LV. Medium 🧐

https://leetcode.com/problems/clone-graph/

Clone Graph - LeetCode

 

문제

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).

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.

 

문제 해결법

Node list를 돌면서 deep copy한 새로운 Node list를 리턴하면 된다.

재귀 함수를 통해 neighbor를 타고 들어가면서 deep copy 해주도록 구현했다.

neighbor을 기준으로 new Node를 만들면 무한루프에 빠질 것이다.

val에 해당하는 Node를 만들었는지 확인해주기 위해 unordered_map를 통해 체크해줬다.

 

해결 코드

class Solution {
private:
	Node *set_neighbor(Node *node, unordered_map<int, Node *> &m){
		Node *new_node = new Node(node->val);
		m[node->val] = new_node;
		for (int i = 0; i < node->neighbors.size(); ++i) {
			if (m.find(node->neighbors[i]->val) != m.end())
				new_node->neighbors.push_back(m[node->neighbors[i]->val]);
			else
				new_node->neighbors.push_back(set_neighbor(node->neighbors[i], m));
		}
		return new_node;
	}
public:
	Node* cloneGraph(Node* node) {
		if (!node)
			return 0;

		// 해당 노드를 추가했나 체크하는 unordered_map
		unordered_map<int, Node *> m;
		return set_neighbor(node, m);
	}
};