[week 18][Leetcode] Clone Graph

• Quetion:

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
         / \
         \_/

• Analysis:

使用广度优先遍历或者深度优先遍历都可以。

• Code:

DFS:

class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(node == nullptr) return nullptr;
        // key is original node value is copied node
        unordered_map<const UndirectedGraphNode *,UndirectedGraphNode *> copied;
        clone(node, copied);
        return copied[node];
    }

private:
    // DFS
    static UndirectedGraphNode* clone(const UndirectedGraphNode *node,unordered_map<const UndirectedGraphNode *,UndirectedGraphNode *> &copied) {
    // a copy already exists
        if (copied.find(node) != copied.end()) return copied[node];
        UndirectedGraphNode *new_node = new UndirectedGraphNode(node->label);
        copied[node] = new_node;
        for (auto nbr : node->neighbors)
            new_node->neighbors.push_back(clone(nbr, copied));
        return new_node;
    }
    
};