【leetcode】clone-graph

写在前面的话：

　　看了看自己的博客，从一月底开始就没怎么更新过，我也确实将近5个月没怎么写代码了。今天突然觉得有些心慌，感觉手都已经生疏了。果然，随便找了道题就卡住了。隐约感觉要用map但又不太记得用法了，知道可以用DFS或BFS却又理不清思路。费了两个小时，结果写了一个shit一样的代码才通过。唉好忧伤啊。

 

 

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/ \\_/

 

我的解法：

反应了好久，才发现题目的难点是防止结点的重复建立。我的方法没有用map，很繁琐。

#include<iostream>
#include<vector>
#include<algorithm>
#include<stdlib.h>
using namespace std;struct UndirectedGraphNode {int label;vector<UndirectedGraphNode *> neighbors;UndirectedGraphNode(int x) : label(x) {};};
class Solution {
public:UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {//获取所有独立的结点vector<UndirectedGraphNode *> UniqueNodes;getUniqueNodes(node, UniqueNodes);return findNode(node, UniqueNodes);}//查找指定结点并返回UndirectedGraphNode * findNode(UndirectedGraphNode * node, vector<UndirectedGraphNode *> UniqueNodes){if(NULL == node)return NULL;for(int i = 0; i < UniqueNodes.size(); i++){if(node->label == UniqueNodes[i]->label){return UniqueNodes[i];}}return NULL;}//获取图中所有的结点和连接信息void getUniqueNodes(UndirectedGraphNode *node, vector<UndirectedGraphNode *>& UniqueNodes){//结点空或已存在时直接返回if(NULL == node || NULL != findNode(node, UniqueNodes))return;//存储新出现的结点UndirectedGraphNode * newNode = new UndirectedGraphNode(node->label);UniqueNodes.push_back(newNode);for(int i = 0; i < node->neighbors.size(); i++){getUniqueNodes(node->neighbors[i], UniqueNodes);newNode->neighbors.push_back(findNode(node->neighbors[i], UniqueNodes));}}
};int main()
{Solution s;UndirectedGraphNode * node0 = new UndirectedGraphNode(0);UndirectedGraphNode * node1 = new UndirectedGraphNode(1);UndirectedGraphNode * node2 = new UndirectedGraphNode(2);node0->neighbors.push_back(node1);node0->neighbors.push_back(node2);node1->neighbors.push_back(node2);node2->neighbors.push_back(node2);UndirectedGraphNode * newNode = s.cloneGraph(node0);return 0;
}

 

网上大神解法

/*** Definition for undirected graph.* struct UndirectedGraphNode {*     int label;*     vector<UndirectedGraphNode *> neighbors;*     UndirectedGraphNode(int x) : label(x) {};* };*/
class Solution {
private:unordered_map<UndirectedGraphNode*,UndirectedGraphNode*> hash;
public://BFSUndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {if(!node) return NULL;queue<UndirectedGraphNode*> Qu;Qu.push(node);hash[node] = new UndirectedGraphNode(node->label);while(!Qu.empty()){UndirectedGraphNode * tmp = Qu.front();Qu.pop();for(UndirectedGraphNode * neighbor : tmp->neighbors){if(hash.find(neighbor) == hash.end()){hash[neighbor] = new UndirectedGraphNode(neighbor->label);Qu.push(neighbor);}hash[tmp]->neighbors.push_back(hash[neighbor]);}}return hash[node];}//DFSUndirectedGraphNode *cloneGraph1(UndirectedGraphNode *node) {if(!node) return NULL;if(hash.find(node) == hash.end()){hash[node] = new UndirectedGraphNode(node->label);for(UndirectedGraphNode* neighbor : node->neighbors){hash[node]->neighbors.push_back(cloneGraph1(neighbor));}}return hash[node];}
};