133. Clone Graph
问题描述:
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 val (int
) and a list (List[Node]
) of its neighbors.
Example:
Input: {"$id":"1","neighbors":[{"$id":"2","neighbors":[{"$ref":"1"},{"$id":"3","neighbors":[{"$ref":"2"},{"$id":"4","neighbors":[{"$ref":"3"},{"$ref":"1"}],"val":4}],"val":3}],"val":2},{"$ref":"4"}],"val":1} Explanation: Node 1's value is 1, and it has two neighbors: Node 2 and 4. Node 2's value is 2, and it has two neighbors: Node 1 and 3. Node 3's value is 3, and it has two neighbors: Node 2 and 4. Node 4's value is 4, and it has two neighbors: Node 1 and 3.
Note:
- The number of nodes will be between 1 and 100.
- The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph.
- Since the graph is undirected, if node p has node q as neighbor, then node q must have node p as neighbor too.
- You must return the copy of the given node as a reference to the cloned graph.
解题思路:
要深度拷贝一个图,每一个节点的值以及与周围点的关系都需要拷贝。
可以用一个map来存储旧节点和新节点的对应关系。
<旧节点,新节点>
对这个图进行遍历,我喜欢使用BFS,用一个queue进行把没有拷贝过的节点放进去。
然后使用一个set存储访问过的节点,防止重复访问。
注意只有在对该节点进行完拷贝后才可以加入到visited set里面去。
代码:
/* // Definition for a Node. class Node { public: int val; vector<Node*> neighbors; Node() {} Node(int _val, vector<Node*> _neighbors) { val = _val; neighbors = _neighbors; } }; */ class Solution { public: Node* cloneGraph(Node* node) { if(node == NULL) return NULL; map<Node*, Node*> record; queue<Node*> nodeQ; set<Node*> visited; nodeQ.push(node); while(!nodeQ.empty()){ Node* cur = nodeQ.front(); if(visited.count(cur) != 0){ nodeQ.pop(); continue; } nodeQ.pop(); if(record.count(cur) == 0){ vector<Node*> neighbors; Node* newNode = new Node(cur->val, neighbors); record[cur] = newNode; } for(auto n : cur->neighbors){ if(visited.count(n) == 0) nodeQ.push(n); if(record.count(n) == 0){ vector<Node*> neighbors; Node* newNode = new Node(n->val, neighbors); record[n] = newNode; } record[cur]->neighbors.push_back(record[n]); } visited.insert(cur); } return record[node]; } };