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Clone Graph.cpp
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Clone Graph.cpp
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/*
Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.
How we serialize an undirected graph:
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.
Link: http://www.lintcode.com/en/problem/clone-graph/
Example:
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
/ \
\_/
Solution: Use a map to find copied root nodes.
Source: https://github.com/kamyu104/LintCode/blob/master/C%2B%2B/clone-graph.cpp
*/
/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
* int label;
* vector<UndirectedGraphNode *> neighbors;
* UndirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
/**
* @param node: A undirected graph node
* @return: A undirected graph node
*/
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
// write your code here
if (!node) {
return nullptr;
}
unordered_map<const UndirectedGraphNode *, UndirectedGraphNode *> copied;
copied[node] = new UndirectedGraphNode(node->label);
queue<const UndirectedGraphNode *> q;
q.emplace(node);
while (!q.empty()) {
auto node = q.front();
q.pop();
for (const auto& n : node->neighbors) {
if (copied.find(n) == copied.end()) {
copied[n] = new UndirectedGraphNode(n->label);
q.emplace(n);
}
copied[node]->neighbors.emplace_back(copied[n]);
}
}
return copied[node];
}
};