133. Clone Graph (3 solutions)——无向无环图复制
Clone Graph
Clone an undirected graph. Each node in the graph contains a label
and a list of its neighbors
.
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 node0
to both nodes1
and2
. - Second node is labeled as
1
. Connect node1
to node2
. - Third node is labeled as
2
. Connect node2
to node2
(itself), thus forming a self-cycle.
Visually, the graph looks like the following:
1
/ \
/ \
0 --- 2
/ \
\_/
这题只需一边遍历一遍复制就可以了。
因此至少可以用三种方法:
1、广度优先遍历(BFS)
2、深度优先遍历(DFS)
2.1、递归
2.2、非递归
解法一:广度优先遍历
变量说明:
映射表m用来保存原图结点与克隆结点的对应关系。
映射表visited用来记录已经访问过的原图结点,防止循环访问。
队列q用于记录广度优先遍历的层次信息。
/**
* 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 == NULL)
return NULL;
// map from origin node to copy node
unordered_map<UndirectedGraphNode *, UndirectedGraphNode *> m;
unordered_map<UndirectedGraphNode *, bool> visited;
queue<UndirectedGraphNode*> q;
q.push(node);
while(!q.empty())
{// BFS
UndirectedGraphNode* front = q.front();
q.pop(); if(visited[front] == false)
{
visited[front] = true; UndirectedGraphNode* cur;
if(m.find(front) == m.end())
{
cur = new UndirectedGraphNode(front->label);
m[front] = cur;
}
else
{
cur = m[front];
}
for(int i = ; i < front->neighbors.size(); i ++)
{
if(m.find(front->neighbors[i]) == m.end())
{
UndirectedGraphNode* nei = new UndirectedGraphNode(front->neighbors[i]->label);
m[front->neighbors[i]] = nei;
cur->neighbors.push_back(nei); q.push(front->neighbors[i]);
}
else
{
cur->neighbors.push_back(m[front->neighbors[i]]);
}
}
}
}
return m[node];
}
};

解法二:递归深度优先遍历(DFS)
/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
* int label;
* vector<UndirectedGraphNode *> neighbors;
* UndirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
map<UndirectedGraphNode *, UndirectedGraphNode *> m; UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node)
{
if(node == NULL)
return NULL; if(m.find(node) != m.end()) //if node is visited, just return the recorded nodeClone
return m[node]; UndirectedGraphNode *nodeClone = new UndirectedGraphNode(node->label);
m[node] = nodeClone;
for(int st = ; st < node->neighbors.size(); st ++)
{
UndirectedGraphNode *temp = cloneGraph(node->neighbors[st]);
if(temp != NULL)
nodeClone->neighbors.push_back(temp);
}
return nodeClone;
}
};
解法三:非递归深度优先遍历(DFS)
深度优先遍历需要进行邻居计数。如果邻居已经全部访问,则该节点访问完成,可以出栈,否则就要继续处理下一个邻居。
/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
* int label;
* vector<UndirectedGraphNode *> neighbors;
* UndirectedGraphNode(int x) : label(x) {};
* };
*/ struct Node
{
UndirectedGraphNode *node;
int ind; //next neighbor to visit
Node(UndirectedGraphNode *n, int i): node(n), ind(i) {}
}; class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
if(node == NULL)
return NULL;
// map from origin node to copy node
unordered_map<UndirectedGraphNode *, UndirectedGraphNode *> m;
unordered_map<UndirectedGraphNode *, bool> visited;
stack<Node*> stk;
Node* newnode = new Node(node, );
stk.push(newnode);
visited[newnode->node] = true;
while(!stk.empty())
{// DFS
Node* top = stk.top();
UndirectedGraphNode* topCopy;
if(m.find(top->node) == m.end())
{
topCopy = new UndirectedGraphNode(top->node->label);
m[top->node] = topCopy;
}
else
topCopy = m[top->node]; if(top->ind == top->node->neighbors.size())
//finished copying its neighbors
stk.pop();
else
{
while(top->ind < top->node->neighbors.size())
{
if(m.find(top->node->neighbors[top->ind]) == m.end())
{
UndirectedGraphNode* neiCopy = new UndirectedGraphNode(top->node->neighbors[top->ind]->label);
m[top->node->neighbors[top->ind]] = neiCopy;
topCopy->neighbors.push_back(neiCopy);
if(visited[top->node->neighbors[top->ind]] == false)
{
visited[top->node->neighbors[top->ind]] = true;
Node* topnei = new Node(top->node->neighbors[top->ind], );
stk.push(topnei);
}
top->ind ++;
break;
}
else
{
topCopy->neighbors.push_back(m[top->node->neighbors[top->ind]]);
top->ind ++;
}
}
}
}
return m[node];
}
};
转自:http://www.cnblogs.com/ganganloveu/p/4119462.html
133. Clone Graph (3 solutions)——无向无环图复制的更多相关文章
- 【LeetCode】133. Clone Graph (3 solutions)
Clone Graph Clone an undirected graph. Each node in the graph contains a label and a list of its nei ...
- 133. Clone Graph 138. Copy List with Random Pointer 拷贝图和链表
133. Clone Graph Clone an undirected graph. Each node in the graph contains a label and a list of it ...
- leetcode 133. Clone Graph ----- java
Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors. OJ's ...
- 133. Clone Graph
题目: Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors. ...
- 133. Clone Graph(图的复制)
Given the head of a graph, return a deep copy (clone) of the graph. Each node in the graph contains ...
- 133. Clone Graph (Graph, Map; DFS)
Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors. OJ's ...
- Graph 133. Clone Graph in three ways(bfs, dfs, bfs(recursive))
Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors. OJ's ...
- Java for LeetCode 133 Clone Graph
Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors. OJ's ...
- [LeetCode] 133. Clone Graph 克隆无向图
Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors. OJ's ...
随机推荐
- PAT Basic 1062
1062 最简分数 一个分数一般写成两个整数相除的形式:N/M,其中 M 不为0.最简分数是指分子和分母没有公约数的分数表示形式. 现给定两个不相等的正分数 N1/M1 和 N2/M ...
- chardet使用方法
简单用法 chardet的使用非常简单,主模块里面只有一个函数detect.detect有一个参数,要求是bytes类型.bytes类型可以通过读取网页内容.open函数的rb模式.带b前缀的字符串. ...
- go 和make的用法 区别
Doand Make are two verbs which frequently confuse students of English. Learn the Difference between ...
- Linux下二进制文件安装MySQL
MySQL 下载地址:https://dev.mysql.com/downloads/mysql/ 并按如下方式选择来下载安装包. 1. 设置配置文件/etc/my.cnmore /etc/my.cn ...
- Python开发:面向对象
Python从设计之初就已经是一门面向对象的语言,正因为如此,在Python中创建一个类和对象是很容易的. 如果你以前没有接触过面向对象的编程语言,那你可能需要先了解一些面向对象语言的一些基本特征,在 ...
- python 四——线程、进程、协程
内容概要 1.进程与线程优.缺点的比较 2.适用情况 3.线程 线程的创建 setDaemon join event RLock 队列 4.进程 创建进程 setDaemon join 线程与进程,数 ...
- NYOJ 722 数独
数独 时间限制:1000 ms | 内存限制:65535 KB 难度:4 描述 数独是一种运用纸.笔进行演算的逻辑游戏.玩家需要根据9×9盘面上的已知数字,推理出所有剩余空格的数字,并满足每一 ...
- DDLog-不同颜色打印信息
(一)下载安装 1.安装插件 XcodeColors Github 链接:https://github.com/robbiehanson/XcodeColors 打开XcodeColors项目,编译即 ...
- 使用Unity做2.5D游戏教程(二)
最近在研究Unity 3D,看了老外Marin Todorov写的教程很详细,就翻译过来以便自己参考,翻译不好的地方请多包涵. 这是使用Unity 游戏开发工具制作一个简单的2.5D 游戏系列教程的第 ...
- [luoguP2051] [AHOI2009]中国象棋(DP)
传送门 注释写明了一切 #include <cstdio> #define N 111 #define p 9999973 #define LL long long int n, m; L ...