[LeetCode] Clone Graph 无向图的复制

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 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
/ \
\_/

这道无向图的复制问题和之前的拷贝带有随机指针的链表有些类似，那道题的难点是如何处理每个节点的随机指针，这道题目的难点在于如何处理每个节点的neighbors，由于在深度拷贝每一个节点后，还要将其所有neighbors放到一个vector中，而如何避免重复拷贝呢？这道题好就好在所有节点值不同，所以我们可以使用哈希表来对应节点值和新生成的节点。对于图的遍历的两大基本方法是深度优先搜索DFS和广度优先搜索BFS，此题的两种解法可参见网友爱做饭的小莹子的博客，这里我们使用深度优先搜索DFS来解答此题，代码如下：

/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
*     int label;
*     vector<UndirectedGraphNode *> neighbors;
*     UndirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
unordered_map<int, UndirectedGraphNode*> umap;
return clone(node, umap);
}
UndirectedGraphNode *clone(UndirectedGraphNode *node, unordered_map<int, UndirectedGraphNode*> &umap) {
if (!node) return node;
if (umap.count(node->label)) return umap[node->label];
UndirectedGraphNode *newNode = new UndirectedGraphNode(node->label);
umap[node->label] = newNode;
for (int i = 0; i < node->neighbors.size(); ++i) {
(newNode->neighbors).push_back(clone(node->neighbors[i], umap));
}
return newNode;
}
};

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