LeetCode 133 Clone Graph （BFS || DFS）

大体题意：

这个题目看了许久 才发现就是一道水题，他就是给你一个无向图，建图的方式都不用你管，他是用vector  保存的！

在给你一个新的点，克隆出这个图来！第一次做leetcode，搞了半天。。

思路：

BFS或者DFS都行 只要能遍历一遍所有的点就好了！

这个图有自环， 加个vis判断是否重复访问即可！

因为是克隆图嘛，可以建立一个unordered_map <UndirectedGraphNode*,UndirectedGraphNode*> mp;

来一个原点和新点的映射！

详细见代码：

BFS版：

class Solution {public:    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {        if (node == NULL)return NULL;        unordered_map <UndirectedGraphNode*,UndirectedGraphNode*> mp;        unordered_map <UndirectedGraphNode*,bool> vis;        queue<UndirectedGraphNode*>q;        mp.clear();        vis.clear();        while(!q.empty()) q.pop();        q.push(node);        vis[node] = false;        while(!q.empty()){            UndirectedGraphNode* u = q.front(); q.pop();            if (vis[u])continue;            vis[u] = true;            if (!mp.count(u)) mp[u] = new UndirectedGraphNode (u->label);            for (int i = 0; i < u->neighbors.size(); ++i){                UndirectedGraphNode* v = u->neighbors[i];                UndirectedGraphNode * tmp;                if (!mp.count(v)){                    tmp = new UndirectedGraphNode (v->label);                    mp[v] = tmp;                }else tmp = mp[v];                q.push(v);                mp[u]->neighbors.push_back(tmp);            }        }        return mp[node];    }};

DFS版：

class Solution {public:    unordered_map<UndirectedGraphNode* ,UndirectedGraphNode* >mp;    unordered_map<UndirectedGraphNode* ,bool >vis;    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {        if (node == NULL)return NULL;        mp.clear();        vis.clear();        dfs(node);        return mp[node];    }    void dfs(UndirectedGraphNode *node){        vis[node] = true;        if (!mp.count(node)) mp[node] = new UndirectedGraphNode(node->label);        UndirectedGraphNode* u = mp[node];        for (int i = 0; i < node->neighbors.size(); ++i){            UndirectedGraphNode *v = node->neighbors[i];            if (vis[v]) continue;            if (!mp.count(v)) mp[v] = new UndirectedGraphNode (v->label);            dfs(v);        }        for (int i = 0; i < node->neighbors.size(); ++i){            UndirectedGraphNode *v = node->neighbors[i];            u->neighbors.push_back(mp[v]);        }    }};

133. Clone Graph

 

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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 #.

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         / \         \_/

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/**

* Definition for undirected graph.

* struct UndirectedGraphNode {

* int label;

* vector<UndirectedGraphNode *> neighbors;

* UndirectedGraphNode(int x) : label(x) {};

* };

*/

classSolution {

public:

UndirectedGraphNode*cloneGraph(UndirectedGraphNode*node){

}

};

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