LeetCode (17) Clone Graph

题目描述

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 #.

First node is labeled as 0. Connect node 0 to both nodes 1 and 2.

Second node is labeled as 1. Connect node 1 to node 2.

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:



本题要求对无向图进行复制。因此本题主要涉及到指针操作，以及深拷贝的概念。从上面的例子图片可以看出，图可能存在自循环，即2号节点同时会指向自己，并且图之间的指向是双向的。因此，我们在实现的时候需要注意哪些节点已经生成了，避免重复操作陷入死循环。题目已经说明，节点的label是唯一的，我们的程序中需要利用本特性。

例如，在生成0号节点的neighbor可以得到1号和2号节点，此时需要生成1号和2号节点。在生成1号节点的时，又会得到其neighbor：0号节点，此时若再生成0，则会陷入死循环。因此，针对label唯一的特性，我们可以使用一个map记录已经生成的node。如果node为生成，则生成新node，若已生成，则将其指针添加到neighbor中即可。

/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
*     int label;
*     vector<UndirectedGraphNode *> neighbors;
*     UndirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
UndirectedGraphNode *clone(UndirectedGraphNode *node, map<int, UndirectedGraphNode*> &labels) {
if (node == NULL)
{
return NULL;
}

if(labels.find(node->label) != labels.end())  //node exist
return labels.find(node->label)->second;
UndirectedGraphNode *g = new UndirectedGraphNode(node->label);
labels[node->label] = g;
for(int i = 0; i < node->neighbors.size(); ++i)
{
g->neighbors.push_back(clone(node->neighbors[i], labels));
}
return g;
}

UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
map<int, UndirectedGraphNode*> labels;
return clone(node, labels);
}
};