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

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

Test case format:

For simplicity sake, each node's value is the same as the node's index (1-indexed). For example, the first node with val = 1, the second node with val = 2, and so on. The graph is represented in the test case using an adjacency list.

Adjacency list is a collection of unordered lists used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.

The given node will always be the first node with val = 1. You must return the copy of the given node as a reference to the cloned graph.

Example 1:

Input: adjList = [[2,4],[1,3],[2,4],[1,3]]
Output: [[2,4],[1,3],[2,4],[1,3]]
Explanation: There are 4 nodes in the graph.
1st node (val = 1)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
2nd node (val = 2)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).
3rd node (val = 3)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
4th node (val = 4)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).

Example 2:

Input: adjList = [[]]
Output: [[]]
Explanation: Note that the input contains one empty list. The graph consists of only one node with val = 1 and it does not have any neighbors.

Example 3:

Input: adjList = []
Output: []
Explanation: This an empty graph, it does not have any nodes.

Example 4:

Input: adjList = [[2],[1]]
Output: [[2],[1]]

实现思路：
本题的题意是克隆一个图，是一道理解C语言指针很好的一道题，别说坑点还是比较多的，实现方法就是深搜遍历，然后创建新的结点，然后遍历原来的图中当前结点的邻结点，互相保存双方的新结点。

AC代码：

class Solution {
		unordered_map<int,int> mp;
		unordered_map<int,Node*> idx;
	public:
		void build(int key) {
			if(idx.count(key)==0) {//当没有出现过该结点的时候需要进行创建
				Node *now=new Node(key);
				idx[key]=now;
			}
		}
		void dfs(Node *root) {
			if(root==nullptr) return;//空结点直接返回
			int k1=root->val;
			build(k1);
			for(int i=0; i<root->neighbors.size(); i++) {
				int k2=root->neighbors[i]->val;
				build(k2);
				if(mp.count(k1*110+k2)==0) {//说明互相的孩子结点中还没有保存
					idx[k1]->neighbors.push_back(idx[k2]);
					idx[k2]->neighbors.push_back(idx[k1]);
					mp[k1*110+k2]=mp[k2*110+k1]=1;
					dfs(root->neighbors[i]);
				}
			}
		}

		Node* cloneGraph(Node* node) {
			dfs(node);
			return idx.size()>0?idx[node->val]:nullptr;
		}
};