LeetCode: 133. Clone Graph

最新推荐文章于 2021-05-14 10:42:19 发布

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最新推荐文章于 2021-05-14 10:42:19 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/qq_26399665/article/details/52624792

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本文介绍了一种无向图的克隆算法，包括广度优先搜索(BFS)和深度优先搜索(DFS)两种实现方式。通过使用队列或递归的方式遍历原图的所有节点并复制它们，同时保持原有的连接关系。

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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:

bfs解决：利用队列，每次遍历队列头结点所有的邻居，如果其某个邻居还没有被创建，那就创建一个，并且将其加入到队列中去．否则说明这个结点已经被创建，而我们只在一个结点刚被创建时才将其加入到队列中去,也就是遍历复制每一个节点，将他和原节点映射在map.

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if (!node)
           return nullptr;
        unordered_map<UndirectedGraphNode*,UndirectedGraphNode*> mp;
        queue<UndirectedGraphNode*> q;
        mp[node]=new UndirectedGraphNode(node->label);
        q.push(node);
        while (!q.empty())
        {
            UndirectedGraphNode* front=q.front();
            q.pop();
            for (int i=0;i<front->neighbors.size();i++)
            {
                if (mp.find(front->neighbors[i])==mp.end())
                {
                    mp[front->neighbors[i]]=new UndirectedGraphNode(front->neighbors[i]->label);
                    q.push(front->neighbors[i]);//front->neighbors[i]);
                }
                mp[front]->neighbors.push_back(mp[front->neighbors[i]]);
            }
        }
        return mp[node];
    }
};

DFS解决:hash表中有就不处理，没有，则复制一个并且遍历它的邻接点，把邻接点递归复制进去。

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if (!node)
           return nullptr;
        if (hash.count(node)) return hash[node];
        hash[node]=new UndirectedGraphNode(node->label);
        for (auto val:node->neighbors)
            hash[node]->neighbors.push_back(cloneGraph(val));
        return hash[node];
    }
private:
 unordered_map<UndirectedGraphNode*,UndirectedGraphNode*> hash;
};