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Copy pathanswer.cpp
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68 lines (60 loc) · 2.48 KB
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/*
Clone Graph Total Accepted: 6392 Total Submissions: 30641 My Submissions
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 nodaes, 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
/ \
\_/
*/
/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
* int label;
* vector<UndirectedGraphNode *> neighbors;
* UndirectedGraphNode(int x) : label(x) {};
* };
*/
/*
这一题考的主要的东西和另外一题有些类似， 就是如何deep copy 一个link， 因为如果直接copy的话copy的是原始的pointer， 因此要找到一个映射， 旧的node和新的node之间的关系
然后再用bfs就可以了
*/
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if(node == NULL) return NULL;
queue<UndirectedGraphNode*> q;
map<int, UndirectedGraphNode*> mp;
UndirectedGraphNode* head = new UndirectedGraphNode(node -> label);
q.push(node);
mp[node -> label] = head;
while(!q.empty()){
UndirectedGraphNode* myNode = q.front();
q.pop();
for(int i = 0; i < myNode -> neighbors.size(); i++){
//has not been visited
if(mp.find(myNode -> neighbors[i] -> label) == mp.end()){
UndirectedGraphNode* newNode = new UndirectedGraphNode(myNode -> neighbors[i] -> label);
mp[myNode -> label] -> neighbors.push_back(newNode);
mp[myNode -> neighbors[i] -> label] = newNode;
q.push(myNode -> neighbors[i]);
}else//has been visited
{
mp[myNode -> label] -> neighbors.push_back(mp[myNode -> neighbors[i] -> label]);
}
}
}
return head;
}
};