Clone Graph

问题来源

问题描述

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

OJ’s undirected graph serialization:
Nodes are labeled uniquely.

We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.
As an example, consider the serialized graph {0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.
   Visually, the graph looks like the following:

       1      / \     /   \    0 --- 2         / \         \_/

 Definition for undirected graph. struct UndirectedGraphNode {     int label;     vector<UndirectedGraphNode *> neighbors;     UndirectedGraphNode(int x) : label(x) {}; };

问题分析

题意大致就是拷贝一个无向图。最开始觉得此题较为简单，我们对传进来的结点进行拷贝，然后对他的邻居结点也做同样处理，显然是一个递归的过程。不过在做了之后我发现，我们还需要做多的一步操作就是，对于已经存在的结点不作二次拷贝。否则，要么我们会在环内进入死循环，要么会在两个相邻结点间建立了重复边。

解决代码

class Solution {public:    map<int, UndirectedGraphNode*> rec; // 记录已生成的结点    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {        if (node == nullptr) return node;        if (rec.find(node->label) == rec.end()) {            rec[node->label] = new UndirectedGraphNode(node->label);            for (auto n : node->neighbors) {                rec[node->label]->neighbors.push_back(cloneGraph(n));            }        }        return rec[node->label];    }};