133. Clone Graph
Medium
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.
Example:
Input:
{"$id":"1","neighbors":[{"$id":"2","neighbors":[{"$ref":"1"},{"$id":"3","neighbors":[{"$ref":"2"},{"$id":"4","neighbors":[{"$ref":"3"},{"$ref":"1"}],"val":4}],"val":3}],"val":2},{"$ref":"4"}],"val":1}
Explanation: Node 1’s value is 1, and it has two neighbors: Node 2 and 4. Node 2’s value is 2, and it has two neighbors: Node 1 and 3. Node 3’s value is 3, and it has two neighbors: Node 2 and 4. Node 4’s value is 4, and it has two neighbors: Node 1 and 3.
Note: The number of nodes will be between 1 and 100. The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph. Since the graph is undirected, if node p has node q as neighbor, then node q must have node p as neighbor too. You must return the copy of the given node as a reference to the cloned graph.
题目大意：给一个图，要求返回一个此图的深复制。
解题思路：图的结构不变，通过一个map保存旧指针到新指针的映射关系。
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/*
// Definition for a Node.
class Node {
public:
int val;
vector<Node*> neighbors;
Node() {}
Node(int _val, vector<Node*> _neighbors) {
val = _val;
neighbors = _neighbors;
}
};
*/
class Solution {
public:
Node* getPtr(unordered_map<Node*, Node*> &mpPtr, Node * const oldPtr){
if(oldPtr == nullptr){
return nullptr;
}
if(mpPtr.count(oldPtr) == 0){
//create the node if it is not yet build
mpPtr[oldPtr] = new Node();
mpPtr[oldPtr]->val = oldPtr->val;
for(auto nh : oldPtr->neighbors){
mpPtr[oldPtr]->neighbors.push_back(getPtr(mpPtr,nh));
}
}
return mpPtr[oldPtr];
}
Node* cloneGraph(Node* node) {
unordered_map<Node*, Node*> mpPtr;
Node *nnode = getPtr(mpPtr, node);
return nnode;
}
};
测试一下，
Success
Details
Runtime: 24 ms, faster than 88.55% of C++ online submissions for Clone Graph.
Memory Usage: 16.7 MB, less than 21.53% of C++ online submissions for Clone Graph.
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