Leetcode Clone Graph

Clone Graph

 Total Accepted: 4360 Total Submissions: 21850My Submissions

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 #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. 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         / \         \_/

这里使用深度优先搜索。这样可以递归实现，如果是宽度优先，就要额外使用queue容器。

关键点：

1 这里的clone需要深度拷贝，就是要使用new操作了

2 防止回路无限循环，就要使用hash表，这里使用unordered_map记录访问过的节点。因为这里的label应该是唯一的才对，所以可以直接使用label作为关键字就可以。

看起来挺难的，因为图总给人困难的感觉，其实不难，3到4星级难度吧，很多都是基本操作组合起来。我一次性通过了。

struct UndirectedGraphNode {int label;vector<UndirectedGraphNode *> neighbors;UndirectedGraphNode(int x) : label(x) {};};class Solution {public:UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {unordered_map<int, UndirectedGraphNode *> track;return cloneGraph(node, track);}UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node, unordered_map<int, UndirectedGraphNode *> &track) {if (!node) return NULL;if (track.count(node->label)) return track[node->label];UndirectedGraphNode *new_node = new UndirectedGraphNode(node->label);new_node->neighbors.resize(node->neighbors.size());track[node->label] = new_node;for (int i = 0; i < node->neighbors.size(); i++){new_node->neighbors[i] = cloneGraph(node->neighbors[i], track);}return new_node;}};

//2014-2-18 updateUndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {unordered_map<int, UndirectedGraphNode *> ump_iu;return clone(node, ump_iu);}UndirectedGraphNode *clone(UndirectedGraphNode *n, unordered_map<int, UndirectedGraphNode *> &ump_iu){if (!n) return n;if (ump_iu.count(n->label)) return ump_iu[n->label];UndirectedGraphNode *rs = new UndirectedGraphNode(n->label);ump_iu[n->label] = rs;for (int i = 0; i < n->neighbors.size(); i++){(rs->neighbors).push_back(clone((n->neighbors[i]), ump_iu));}return rs;}