【Clone Graph】cpp
题目:
Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.
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
/ \
\_/
代码:
/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
* int label;
* vector<UndirectedGraphNode *> neighbors;
* UndirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node)
{
map<UndirectedGraphNode *, UndirectedGraphNode *> copied;
return Solution::dfs(copied, node);
}
static UndirectedGraphNode *dfs(
map<UndirectedGraphNode *, UndirectedGraphNode *>& copied,
UndirectedGraphNode *node)
{
if ( node == NULL) return NULL;
if ( copied.find(node)!=copied.end() ) return copied[node];
UndirectedGraphNode *cloneNode = new UndirectedGraphNode(node->label);
copied[node] = cloneNode;
for ( int i=; i<node->neighbors.size(); ++i )
{
cloneNode->neighbors.push_back( Solution::dfs(copied,node->neighbors[i]) );
}
return cloneNode;
}
};
tips:
图的深拷贝。
学到的一个技巧是如何不重复拷贝图中的node:用一个map<node *, node *>记录已经拷贝过的原图中的点以及其对应的新图中的点。
剩下的就是按照深搜模板来完成。
====================================
第二次过这道题,照着之前的思路写,漏掉了重要的红字的部分。
/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
* int label;
* vector<UndirectedGraphNode *> neighbors;
* UndirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node)
{
map<UndirectedGraphNode*, UndirectedGraphNode*> originCopy;
return Solution::dfs(node, originCopy);
}
static UndirectedGraphNode* dfs(
UndirectedGraphNode* origin,
map<UndirectedGraphNode*, UndirectedGraphNode*>& originCopy)
{
if ( !origin ) return NULL;
if ( originCopy.find(origin)!=originCopy.end() ) return originCopy[origin];
UndirectedGraphNode* copy = new UndirectedGraphNode(origin->label);
originCopy[origin] = copy;
for ( int i=; i<origin->neighbors.size(); ++i )
{
copy->neighbors.push_back(Solution::dfs(origin->neighbors[i], originCopy));
}
return copy;
}
};
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