Today I have done leetcode's February LeetCoding Challenge with cpp.

February LeetCoding Challenge 23

Description

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).

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.

Solution

class Solution {
  map<int, Node*> nodes;
public:
  Node* cloneGraph(Node* node) {
    if(!node) return node;
    Node *newNode = new Node(node->val);
    nodes[node->val] = newNode;
    for(auto neighbor : node->neighbors) {
      if(nodes.count(neighbor->val)) {
        newNode->neighbors.push_back(nodes[neighbor->val]);
      } else {
        newNode->neighbors.push_back(cloneGraph(neighbor));
      }
    }
    return newNode;
  }
};

// Accepted
// 22/22 cases passed (15 ms)
// Your runtime beats 14.73 % of cpp submissions
// Your memory usage beats 59.68 % of cpp submissions (8.7 MB)