数学基础

The definition of a graph should just involve sets(set of vertices and set of edges).

入门题

133. Clone Graph

视频解析

解题思路

因为题目已经说清楚为Connected graph, 所以我们可以先过一遍所有的点，将点进行拷贝，再将每个点的edge set进行拷贝。为防止重复拷贝以及方便寻找，用unordered_map<Node*,Node*> 记录原graph的点和复制后graph的点。

Time Complexity O(M+N) M: number of nodes, N : number of edges

Space Complexity O(M) M: number of nodes

class Solution {
    unordered_map<Node*, Node*> m;
public:
    Node* cloneGraph(Node* node) {
        if(!node) return nullptr;
        dfs(node);
        for(auto [from, to]: m){
            for(auto n: from->neighbors){
                to->neighbors.push_back(m[n]);
            }
        }
        return m[node];
    }
    void dfs(Node* node){
        m[node] = new Node(node->val);
        for(auto next: node->neighbors){
            if(!m[next]){
                dfs(next);
            }
        }
    }
};

类似题

• 138. Copy List with Random Pointer (Medium)
• 1490. Clone N-ary Tree (Medium)
• 1485. Clone Binary Tree With Random Pointer

Clone/copy With Random Pointer 是最像的，思路一模一样。Clone N-ary Tree可以使用同样的也可以不单独列dfs function 直接callback本身copy_n->children.push_back(cloneTree(child));Time/Space Complexity 一致