leetcode Clone Graph

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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//首先，这种深拷贝的题目，需要有一个辅助工具来判断中的节点是否重复。本题是图的拷贝，按照某一个规律全部把图的节点全部遍历一遍。

//我是用map来判断是否重复，然后根据原图进行广度优先遍历，然后把新图的节点一一new出来，并对节点进行相应的连接。

/** * Definition for undirected graph. * struct UndirectedGraphNode { *     int label; *     vector<UndirectedGraphNode *> neighbors; *     UndirectedGraphNode(int x) : label(x) {}; * }; */ //7:35->class Solution {public:    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {        map<UndirectedGraphNode *,UndirectedGraphNode *> repeat;        // one set and two queue        // store the original nodes and new nodes;que is for old graph. que2 is for new graph.        queue<UndirectedGraphNode *> que;        queue<UndirectedGraphNode *> que2;        UndirectedGraphNode *head=NULL,*h1=NULL,*h2=NULL,*h3=NULL;        int i=0,j=0;        if(node==NULL)        {            return head;        }        que.push(node);        h1=head = new UndirectedGraphNode(node->label);        que2.push(h1);        repeat[node] = h1;        // begin bfs        while(que.size()>0)        {            node = que.front();            que.pop();            h1 = que2.front();            que2.pop();            for(i=0;i<node->neighbors.size();i++)            {                h2 = node->neighbors[i];                if(repeat.count(h2)==0)                {                    h3 = new UndirectedGraphNode(h2->label);                    h1->neighbors.push_back(h3);                    repeat[h2] = h3;                    que.push(h2);                    que2.push(h3);                }else                {//add the new link in new graph                    h1->neighbors.push_back(repeat[h2]);                }            }        }        return head;    }};