0%

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

题目描述

题目链接:133. Clone Graph

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:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
#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