133. Clone Graph（DFS&BFS）

1.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:

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

2. 代码

2.1

1.使用到的知识：unordered_map

2.参考链接：点击

2.2 DFS实现

/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
*     int label;
*     vector<UndirectedGraphNode *> neighbors;
*     UndirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
if(node == NULL) return node;
if(mmp.count(node->label) > 0)
return mmp[node->label];
UndirectedGraphNode* newNode = new UndirectedGraphNode(node->label);
mmp[node->label] = newNode;
for(int i = 0; i < node->neighbors.size(); i++) {
(newNode->neighbors).push_back(cloneGraph(node->neighbors[i]));
}
return newNode;
}

private:
unordered_map<int, UndirectedGraphNode*> mmp;
};

2.3 BFS实现

写在前面：

1. 这道题应该是用DFS好一点。

2. 下面的BFS解法涉及一点“迭代器”的知识，其实可以用简单的下标访问代替。

UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
// BFS copy.
if (node == nullptr) {
return nullptr;
}
queue<UndirectedGraphNode *> BFSQueue;
BFSQueue.push(node);
unordered_map<int, UndirectedGraphNode *> visited;
// Copy the first node.
UndirectedGraphNode *copy = new UndirectedGraphNode(node->label);
// Mark as visited.
visited[node->label] = copy;
while (!BFSQueue.empty()) {
UndirectedGraphNode *oNode = BFSQueue.front();
BFSQueue.pop();
// Copy the list of its neighbors.
vector<UndirectedGraphNode *> oNeighbors = oNode->neighbors;
for (vector<UndirectedGraphNode *>::const_iterator it = oNeighbors.begin(); it != oNeighbors.end(); ++it) {
if (visited.count((*it)->label) == 0) {
UndirectedGraphNode *neighborNode = new UndirectedGraphNode((*it)->label);
visited[(*it)->label] = neighborNode;
BFSQueue.push(*it);
}
(visited[oNode->label]->neighbors).push_back(visited[(*it)->label]);
}
}
return copy;
}

部分参考leetcode的朋友：参考

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