首页 > 代码库 > [Leetcode]Clone Graph

[Leetcode]Clone Graph

2024-08-05 20:42:42   218人阅读

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
         /          \_/

Hide Tags
 Depth-first Search Breadth-first Search Graph



题目要求复制一个图
根据题意我们发现lable是独一无二的，它标志了一个node。因此我们可以用关联容器map，用lable作为KEY来保存已经创建过的点的地址
先写了一个递归的版本：

class Solution {public:<span style="white-space:pre">	</span>UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {<span style="white-space:pre">		</span>if (NULL == node) return NULL;<span style="white-space:pre">		</span>map<int, UndirectedGraphNode *> occured;<span style="white-space:pre">		</span>return clone(node, occured);<span style="white-space:pre">	</span>}<span style="white-space:pre">	</span>UndirectedGraphNode *clone(UndirectedGraphNode *node, map<int, UndirectedGraphNode *> &occured){<span style="white-space:pre">		</span>if (occured.count(node->label) == 0){<span style="white-space:pre">			</span>UndirectedGraphNode *newNode = new UndirectedGraphNode(node->label);<span style="white-space:pre">			</span>occured[node->label] = newNode;<span style="white-space:pre">			</span>for (int i = 0; i < node->neighbors.size(); i++){<span style="white-space:pre">				</span>newNode->neighbors.push_back(clone(node->neighbors[i], occured));<span style="white-space:pre">			</span>}<span style="white-space:pre">			</span>return newNode;<span style="white-space:pre">		</span>}<span style="white-space:pre">		</span>else{<span style="white-space:pre">			</span>return occured[node->label];<span style="white-space:pre">		</span>}<span style="white-space:pre">	</span>}};

然后又写了一个非递归版本，这一次采用每个节点在原图中的地址作为KEY，来储存对应新节点的地址
用一个queue来保存要访问的节点

class Solution {
public:
	UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
		if (NULL == node) return NULL;
		map<UndirectedGraphNode *, UndirectedGraphNode *> occured;
		queue<UndirectedGraphNode *> visit;
		visit.push(node);
		while (!visit.empty()){
			UndirectedGraphNode *cur = visit.front();
			visit.pop();
			if (occured.count(cur) == 0){
				UndirectedGraphNode *newNode = new UndirectedGraphNode(cur->label);
				occured[cur] = newNode;
			}
			for (int i = 0; i < cur->neighbors.size(); i++){
				UndirectedGraphNode *tmp = cur->neighbors[i];
				if (occured.count(tmp) == 0){
					UndirectedGraphNode *newNode = new UndirectedGraphNode(tmp->label);
					occured[tmp] = newNode;
					visit.push(tmp);
				}
				occured[cur]->neighbors.push_back(occured[tmp]);
			}
		}
		return occured[node];
	}
};

两段代码都是一遍AC

[Leetcode]Clone Graph


联系
我们