Clone Graph Problem
Description
LeetCode Problem 133.
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.
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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:
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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:
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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:
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Input: adjList = []
Output: []
Explanation: This an empty graph, it does not have any nodes.
Example 4:
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Input: adjList = [[2],[1]]
Output: [[2],[1]]
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.
Sample C++ Code
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class Solution {
public:
Node* cloneGraph(Node* node) {
queue<Node*> q, qc;
if (node == NULL)
return NULL;
Node* nodec = new Node(node->val);
q.push(node);
qc.push(nodec);
unordered_map<Node*, Node*> cloned;
cloned[node] = nodec;
Node* n;
Node* nc;
while (!q.empty()) {
n = q.front();
q.pop();
nc = qc.front();
qc.pop();
for (auto j : n->neighbors) {
if (j == NULL)
continue;
if (cloned.find(j) == cloned.end()) {
Node* jc = new Node(j->val);
nc->neighbors.push_back(jc);
q.push(j);
qc.push(jc);
cloned[j] = jc;
} else {
nc->neighbors.push_back(cloned[j]);
}
}
}
return nodec;
}
};