[Leetcode]133. Clone Graph(C++)

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 value (int) and a list (List[Node]) of its neighbors.

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

例子

见上方链接。

Constraints

The number of nodes in the graph is in the range [0, 100].

1 <= Node.val <= 100

Node.val is unique for each node.

There are 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.

解题思路

用广度优先搜索进行原图的遍历，遍历过程中用哈希表来维护两个图中对应的节点。遇到不在哈希表中的节点时则新建节点，在遍历邻居过程中完成新图中的节点链接关系。代码如下：

#include <queue>
#include <unordered_map>
#include <unordered_set>
class Solution {
public:
    Node* cloneGraph(Node* node) {
        if (node == nullptr) {
            return nullptr;
        }
        Node* ret = new Node(node->val);

        std::unordered_map<Node*, Node*> hmap;
        hmap[node] = ret;
        std::unordered_set<Node*> visited{node};

        // bfs
        std::queue<Node*> current_level;
        current_level.push(node);
        while (!current_level.empty()) {
            std::queue<Node*> next_level;
            while (!current_level.empty()) {
                Node* n = current_level.front();
                current_level.pop();

                for (auto nei : n->neighbors) {
                    if (hmap.find(nei) == hmap.end()) {
                        Node* copy = new Node(nei->val);
                        hmap[nei] = copy;
                    }
                    hmap[n]->neighbors.push_back(hmap[nei]);

                    if (visited.count(nei) == 0) {
                        next_level.push(nei);
                        visited.insert(nei);
                    }
                }
            }
            std::swap(next_level, current_level);
        }
        return ret;
    }
};

时间复杂度: O(|V|+|E|)
空间复杂度: O(|V|+|E|)

GitHub 代码同步地址： 133.CloneGraph.cpp

其他题目： GitHub: Leetcode-C++-Solution 博客： Leetcode-Solutions