[LeetCode] 133. Clone Graph 克隆無向圖

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:

1. The number of nodes will be between 1 and 100.
2. The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph.
3. Since the graph is undirected, if node p has node q as neighbor, then node q must have node p as neighbor too.
4. You must return the copy of the given node as a reference to the cloned graph.

這道無向圖的復(fù)制問題和之前的 Copy List with Random Pointer 有些類似，那道題的難點(diǎn)是如何處理每個(gè)結(jié)點(diǎn)的隨機(jī)指針，這道題目的難點(diǎn)在于如何處理每個(gè)結(jié)點(diǎn)的 neighbors，由于在深度拷貝每一個(gè)結(jié)點(diǎn)后，還要將其所有 neighbors 放到一個(gè) vector 中，而如何避免重復(fù)拷貝呢？這道題好就好在所有結(jié)點(diǎn)值不同，所以我們可以使用 HashMap 來對(duì)應(yīng)原圖中的結(jié)點(diǎn)和新生成的克隆圖中的結(jié)點(diǎn)。對(duì)于圖的遍歷的兩大基本方法是深度優(yōu)先搜索 DFS 和廣度優(yōu)先搜索 BFS，這里我們先使用深度優(yōu)先搜索DFS來解答此題，在遞歸函數(shù)中，首先判空，然后再看當(dāng)前的結(jié)點(diǎn)是否已經(jīng)被克隆過了，若在 HashMap 中存在，則直接返回其映射結(jié)點(diǎn)。否則就克隆當(dāng)前結(jié)點(diǎn)，并在 HashMap 中建立映射，然后遍歷當(dāng)前結(jié)點(diǎn)的所有 neihbor 結(jié)點(diǎn)，調(diào)用遞歸函數(shù)并且加到克隆結(jié)點(diǎn)的 neighbors 數(shù)組中即可，代碼如下：

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;
    }
};

我們也可以使用 BFS 來遍歷圖，使用隊(duì)列 queue 進(jìn)行輔助，還是需要一個(gè) HashMap 來建立原圖結(jié)點(diǎn)和克隆結(jié)點(diǎn)之間的映射。先克隆當(dāng)前結(jié)點(diǎn)，然后建立映射，并加入 queue 中，進(jìn)行 while 循環(huán)。在循環(huán)中，取出隊(duì)首結(jié)點(diǎn)，遍歷其所有 neighbor 結(jié)點(diǎn)，若不在 HashMap 中，我們根據(jù) neigbor 結(jié)點(diǎn)值克隆一個(gè)新 neighbor 結(jié)點(diǎn)，建立映射，并且排入 queue 中。然后將 neighbor 結(jié)點(diǎn)在 HashMap 中的映射結(jié)點(diǎn)加入到克隆結(jié)點(diǎn)的 neighbors 數(shù)組中即可，參見代碼如下：

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;
    }
};

類似題目：

Copy List with Random Pointer

參考資料：

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

https://leetcode.com/problems/clone-graph/discuss/42313/C%2B%2B-BFSDFS

https://leetcode.com/problems/clone-graph/discuss/42309/Depth-First-Simple-Java-Solution

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