[LeetCode] Clone Graph
Clone Graph
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 node0to both nodes1and2.Second node is labeled as
1. Connect node1to node2.Third node is labeled as
2. Connect node2to node2(itself), thus forming a self-cycle.
Visually, the graph looks like the following:
1 / \ / \ 0 --- 2 / \ \_/
解题思路：
题意为拷贝图。我们可以用一个map来存储新图的标签到节点指针的映射关系。同时，用一个队列来存储待确定邻居的节点（原图节点指针，因为我们通过指针标签很容易找到新图节点指针）。
/**
* 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 NULL;
}
map<int, UndirectedGraphNode*> nodeMap; //存的是新图的标签到节点指针对
queue<UndirectedGraphNode*> nodeQ; //存的是原图的节点指针
UndirectedGraphNode* newNode = new UndirectedGraphNode(node->label);
nodeMap[node->label] = newNode;
nodeQ.push(node);
while(!nodeQ.empty()){
UndirectedGraphNode* n = nodeQ.front();
nodeQ.pop();
for(int i=0; i<n->neighbors.size(); i++){
if(nodeMap.find(n->neighbors[i]->label) == nodeMap.end()){
nodeMap[n->neighbors[i]->label] = new UndirectedGraphNode(n->neighbors[i]->label);
nodeQ.push(n->neighbors[i]);
}
nodeMap[n->label]->neighbors.push_back(nodeMap[n->neighbors[i]->label]);
}
}
return newNode;
}
};
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