【Clone Graph】cpp
题目:
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 node0to both nodes1and2. - Second node is labeled as
1. Connect node1to node2. - Third node is labeled as
2. Connect node2to node2(itself), thus forming a self-cycle.
Visually, the graph looks like the following:
1
/ \
/ \
0 --- 2
/ \
\_/
代码:
/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
* int label;
* vector<UndirectedGraphNode *> neighbors;
* UndirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node)
{
map<UndirectedGraphNode *, UndirectedGraphNode *> copied;
return Solution::dfs(copied, node);
}
static UndirectedGraphNode *dfs(
map<UndirectedGraphNode *, UndirectedGraphNode *>& copied,
UndirectedGraphNode *node)
{
if ( node == NULL) return NULL;
if ( copied.find(node)!=copied.end() ) return copied[node];
UndirectedGraphNode *cloneNode = new UndirectedGraphNode(node->label);
copied[node] = cloneNode;
for ( int i=; i<node->neighbors.size(); ++i )
{
cloneNode->neighbors.push_back( Solution::dfs(copied,node->neighbors[i]) );
}
return cloneNode;
}
};
tips:
图的深拷贝。
学到的一个技巧是如何不重复拷贝图中的node:用一个map<node *, node *>记录已经拷贝过的原图中的点以及其对应的新图中的点。
剩下的就是按照深搜模板来完成。
====================================
第二次过这道题,照着之前的思路写,漏掉了重要的红字的部分。
/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
* int label;
* vector<UndirectedGraphNode *> neighbors;
* UndirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node)
{
map<UndirectedGraphNode*, UndirectedGraphNode*> originCopy;
return Solution::dfs(node, originCopy);
}
static UndirectedGraphNode* dfs(
UndirectedGraphNode* origin,
map<UndirectedGraphNode*, UndirectedGraphNode*>& originCopy)
{
if ( !origin ) return NULL;
if ( originCopy.find(origin)!=originCopy.end() ) return originCopy[origin];
UndirectedGraphNode* copy = new UndirectedGraphNode(origin->label);
originCopy[origin] = copy;
for ( int i=; i<origin->neighbors.size(); ++i )
{
copy->neighbors.push_back(Solution::dfs(origin->neighbors[i], originCopy));
}
return copy;
}
};
【Clone Graph】cpp的更多相关文章
- hdu 4739【位运算】.cpp
题意: 给出n个地雷所在位置,正好能够组成正方形的地雷就可以拿走..为了简化题目,只考虑平行于横轴的正方形.. 问最多可以拿走多少个正方形.. 思路: 先找出可以组成正方形的地雷组合cnt个.. 然后 ...
- Hdu 4734 【数位DP】.cpp
题意: 我们定义十进制数x的权值为f(x) = a(n)*2^(n-1)+a(n-1)*2(n-2)+...a(2)*2+a(1)*1,a(i)表示十进制数x中第i位的数字. 题目给出a,b,求出0~ ...
- 【Valid Sudoku】cpp
题目: Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku board could ...
- 【Permutations II】cpp
题目: Given a collection of numbers that might contain duplicates, return all possible unique permutat ...
- 【Subsets II】cpp
题目: Given a collection of integers that might contain duplicates, nums, return all possible subsets. ...
- 【Sort Colors】cpp
题目: Given an array with n objects colored red, white or blue, sort them so that objects of the same ...
- 【Sort List】cpp
题目: Sort a linked list in O(n log n) time using constant space complexity. 代码: /** * Definition for ...
- 【Path Sum】cpp
题目: Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up ...
- 【Symmetric Tree】cpp
题目: Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). F ...
随机推荐
- SpringBoot的优缺点
优点: 1.快速构建项目 2.对主流开发框架的无配置继承 3.项目可独立运行,无须外部依赖Servlet容器 4.提高运行时的应用监控 5.极大地提高了开发.部署效率 6.与云计算的天然集成 缺点: ...
- 使用startForeground让android服务前台运行
最近在使用android 4.1系统的时候,发现在手机休眠一段时间后(1-2小时),后台运行的服务被强行kill掉,有可能是系统回收内存的一种机制,要想避免这种情况可以通过startForegroun ...
- JSP开发过程遇到的中文乱码问题及解决方案
对于程序猿来说,乱码问题真的很头疼,下面列举几种常见的乱码. 1.数据库编码不一致导致乱码 解决方法: 首先查看数据库编码,输入: show variables like "%char%&q ...
- 动态规划专题(五)——斜率优化DP
前言 斜率优化\(DP\)是难倒我很久的一个算法,我花了很长时间都难以理解.后来,经过无数次的研究加以对一些例题的理解,总算啃下了这根硬骨头. 基本式子 斜率优化\(DP\)的式子略有些复杂,大致可以 ...
- centos 7 虚拟机启用网卡
1.vi /etc/sysconfig/network-scripts/ifcfg-enp0s3 2.编辑默认网卡配置文件,将ONBOOT由no改为yes,编辑完成后,按ESC回至命令模板,输入&qu ...
- Jmeter命令行参数
一.在linux中,使用非gui的方式执行jmeter.若需更改参数,必须先编辑jmx文件,找到对应的变量进行修改,比较麻烦.因此,可以参数化一些常用的变量,直接在Jmeter命令行进行设置 二.参数 ...
- 记Tea使用中遇到的问题及解决过程
学习Markdown时,在小众软件看到一个叫Tea的软件.UI设计是简约风格:"所见即所得"的Markdown:支持插件等原因让我选择去尝试这杯"茶". 最近一 ...
- 基于CXF开发crm服务
1 基于CXF开发crm服务 1.1 数据库环境搭建 1.2 web项目环境搭建 第一步:创建动态web项目 第二步:导入CXF相关jar包 第三步:配置web.xml <context-par ...
- 谭浩强 c++程序设计第一章课后习题 第7题
#include <iostream> using namespace std; int main() { int a,b,c; int f(int x,int y,int z);//这是 ...
- Ubuntu 14.04 LTS 触摸板无法使用
c16b上,触摸板不能使用,查找后发现,需要在加载驱动时增加参数. 如下所说: 1.使用以下命令后,触摸板可以使用 sudo modprobe -r psmouse sudo modprobe psm ...