133. Clone Graph

Difficulty: Medium

Frequency: N/A

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

OJ’s undirected graph serialization:Nodes are labeled uniquely.

We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.As an example, consider the serialized graph {0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by #.

  1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
  2. Second node is labeled as 1. Connect node 1 to node 2.
  3. Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

Visually, the graph looks like the following:

       1
      / \
     /   \
    0 --- 2
         / \
         \_/

Test Cases:


Solution 1:

Data structure:
Steps:
Complexity:
Runtime:
Space:
Code:
/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if (!node) return NULL;
        UndirectedGraphNode *p1 = node;
        UndirectedGraphNode *p2 = new UndirectedGraphNode(node->label);
        unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> m;
        queue<UndirectedGraphNode*> q;
        q.push(node);
        m[node] = p2;
        while (!q.empty()) {
            p1 = q.front();
            p2 = m[p1];
            q.pop();
            for (int i = 0; i < p1->neighbors.size(); i++) {
                UndirectedGraphNode *nb = p1->neighbors[i];
                if (m.count(nb)) {
                    p2->neighbors.push_back(m[nb]);
                } else {
                    UndirectedGraphNode *tmp = new UndirectedGraphNode(nb->label);
                    p2->neighbors.push_back(tmp);
                    m[nb] = tmp;
                    q.push(nb);
                }
            }
        }
        return m[node];
    }
};

Solution 2:

Data structure:
steps:
Complexity:
Runtime:
Space:
Code:


Submission errors:

 

Things to learn:

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