题目地址

https://leetcode.com/problems/clone-graph/

题目描述

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.

class Node {
    public int val;
    public List<Node> neighbors;
}

Test case format:

For simplicity sake, 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.

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.

Example 4:

Input: adjList = [[2],[1]]
Output: [[2],[1]]

Constraints:

• 1 <= Node.val <= 100
• Node.val is unique for each node.
• Number of Nodes will not exceed 100.
• There is 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.

思路

这道无向图的复制问题和之前的 Copy List with Random Pointer 有些类似，那道题的难点是如何处理每个结点的随机指针，这道题目的难点在于如何处理每个结点的 neighbors，由于在深度拷贝每一个结点后，还要将其所有 neighbors 放到一个 vector 中，而如何避免重复拷贝呢？这道题好就好在所有结点值不同，所以我们可以使用 HashMap 来对应原图中的结点和新生成的克隆图中的结点。对于图的遍历的两大基本方法是深度优先搜索 DFS 和广度优先搜索 BFS，这里我们先使用深度优先搜索DFS来解答此题，在递归函数中，首先判空，然后再看当前的结点是否已经被克隆过了，若在 HashMap 中存在，则直接返回其映射结点。否则就克隆当前结点，并在 HashMap 中建立映射，然后遍历当前结点的所有 neihbor 结点，调用递归函数并且加到克隆结点的 neighbors 数组中即可

关键点解析

代码

解法一：

class Solution {
public:
    Node* cloneGraph(Node* node) {
        unordered_map<Node*, Node*> m;
        return helper(node, m);
    }
    Node* helper(Node* node, unordered_map<Node*, Node*>& m) {
        if (!node) return NULL;
        if (m.count(node)) return m[node];
        Node *clone = new Node(node->val);
        m[node] = clone;
        for (Node *neighbor : node->neighbors) {
            clone->neighbors.push_back(helper(neighbor, m));
        }
        return clone;
    }
};

我们也可以使用 BFS 来遍历图，使用队列 queue 进行辅助，还是需要一个 HashMap 来建立原图结点和克隆结点之间的映射。先克隆当前结点，然后建立映射，并加入 queue 中，进行 while 循环。在循环中，取出队首结点，遍历其所有 neighbor 结点，若不在 HashMap 中，我们根据 neigbor 结点值克隆一个新 neighbor 结点，建立映射，并且排入 queue 中。然后将 neighbor 结点在 HashMap 中的映射结点加入到克隆结点的 neighbors 数组中即可，参见代码如下：

解法二：

class Solution {
public:
    Node* cloneGraph(Node* node) {
        if (!node) return NULL;
        unordered_map<Node*, Node*> m;
        queue<Node*> q{{node}};
        Node *clone = new Node(node->val);
        m[node] = clone;
        while (!q.empty()) {
            Node *t = q.front(); q.pop();
            for (Node *neighbor : t->neighbors) {
                if (!m.count(neighbor)) {
                    m[neighbor] = new Node(neighbor->val);
                    q.push(neighbor);
                }
                m[t]->neighbors.push_back(m[neighbor]);
            }
        }
        return clone;
    }
};

本文参考自：
https://github.com/grandyang/leetcode/ &
https://github.com/azl397985856/leetcode