Given a list of directory info including directory path, and all the files with contents in this directory, you need to find out all the groups of duplicate files in the file system in terms of their paths.

A group of duplicate files consists of at least two files that have exactly the same content.

A single directory info string in the input list has the following format:

"root/d1/d2/.../dm f1.txt(f1_content) f2.txt(f2_content) ... fn.txt(fn_content)"

It means there are n files (f1.txtf2.txt ... fn.txt with content f1_contentf2_content ... fn_content, respectively) in directory root/d1/d2/.../dm. Note that n >= 1 and m >= 0. If m = 0, it means the directory is just the root directory.

The output is a list of group of duplicate file paths. For each group, it contains all the file paths of the files that have the same content. A file path is a string that has the following format:

"directory_path/file_name.txt"

Example 1:

Input:
["root/a 1.txt(abcd) 2.txt(efgh)", "root/c 3.txt(abcd)", "root/c/d 4.txt(efgh)", "root 4.txt(efgh)"]
Output:
[["root/a/2.txt","root/c/d/4.txt","root/4.txt"],["root/a/1.txt","root/c/3.txt"]]

Note:

  1. No order is required for the final output.
  2. You may assume the directory name, file name and file content only has letters and digits, and the length of file content is in the range of [1,50].
  3. The number of files given is in the range of [1,20000].
  4. You may assume no files or directories share the same name in the same directory.
  5. You may assume each given directory info represents a unique directory. Directory path and file info are separated by a single blank space.

Follow-up beyond contest:

  1. Imagine you are given a real file system, how will you search files? DFS or BFS?
  2. If the file content is very large (GB level), how will you modify your solution?
  3. If you can only read the file by 1kb each time, how will you modify your solution?
  4. What is the time complexity of your modified solution? What is the most time-consuming part and memory consuming part of it? How to optimize?
  5. How to make sure the duplicated files you find are not false positive?
 Approach #1: C++.
class Solution {
public:
vector<vector<string>> findDuplicate(vector<string>& paths) {
int size = paths.size();
unordered_map<string, vector<string>> mp;
for (int i = 0; i < size; ++i) {
int found = paths[i].find(' ');
string str = paths[i].substr(0, found);
while (found != string::npos) {
int last = found;
found = paths[i].find(' ', last+1);
mp[str].push_back(paths[i].substr(last+1, found-last-1));
}
} unordered_map<string, vector<string>> temp; for (auto m : mp) {
string base = m.first + "/";
for (string s : m.second) {
int fre = s.find('(');
int las = s.find(')');
string key = s.substr(fre+1, las-fre-1);
string kid = s.substr(0, fre);
temp[key].push_back(base+kid);
}
} vector<vector<string>> ans; for (auto it : temp) {
vector<string> ant;
for (string s : it.second) {
ant.push_back(s);
}
if (ant.size() >1)
ans.push_back(ant);
} return ans;
}
};

  

Approach #2: Java.

class Solution {
public List<List<String>> findDuplicate(String[] paths) {
HashMap<String, List<String>> map = new HashMap<>();
for (String path : paths) {
String[] values = path.split(" ");
for (int i = 1; i < values.length; ++i) {
String[] name_cont = values[i].split("\\(");
name_cont[1] = name_cont[1].replace(")", "");
List<String> list = map.getOrDefault(name_cont[1], new ArrayList<String>());
list.add(values[0] + "/" + name_cont[0]);
map.put(name_cont[1], list);
}
}
List<List<String>> res = new ArrayList<>();
for (String key : map.keySet()) {
if (map.get(key).size() > 1)
res.add(map.get(key));
}
return res;
}
}

  

Apparoch #3: Python.

class Solution(object):
def findDuplicate(self, paths):
"""
:type paths: List[str]
:rtype: List[List[str]]
"""
M = collections.defaultdict(list)
for line in paths:
data = line.split()
root = data[0]
for file in data[1:]:
name, _, content = file.partition('(')
M[content[:-1]].append(root + '/' + name)
return [x for x in M.values() if len(x) > 1]

  

Analysis:

In this question our goal is to split and combine the string. If you are familiar with the operate it will easy to solve this problem.

C++ -----> string:assign

string (1)
string& assign (const string& str);
substring (2)
string& assign (const string& str, size_t subpos, size_t sublen);
c-string (3)
string& assign (const char* s);
buffer (4)
string& assign (const char* s, size_t n);
fill (5)
string& assign (size_t n, char c);
range (6)
template <class InputIterator>
string& assign (InputIterator first, InputIterator last);
Assign content to string

Assigns a new value to the string, replacing its current contents.

(1) string
Copies str.
(2) substring
Copies the portion of str that begins at the character position subpos and spans sublen characters (or until the end of str, if either str is too short or if sublen is string::npos).
(3) c-string
Copies the null-terminated character sequence (C-string) pointed by s.
(4) buffer
Copies the first n characters from the array of characters pointed by s.
(5) fill
Replaces the current value by n consecutive copies of character c.
(6) range
Copies the sequence of characters in the range [first,last), in the same order.
(7) initializer list
Copies each of the characters in il, in the same order.
(8) move
Acquires the contents of str.
str is left in an unspecified but valid state.
// string::assign
#include <iostream>
#include <string> int main ()
{
std::string str;
std::string base="The quick brown fox jumps over a lazy dog."; // used in the same order as described above: str.assign(base);
std::cout << str << '\n'; str.assign(base,10,9);
std::cout << str << '\n'; // "brown fox" str.assign("pangrams are cool",7);
std::cout << str << '\n'; // "pangram" str.assign("c-string");
std::cout << str << '\n'; // "c-string" str.assign(10,'*');
std::cout << str << '\n'; // "**********" str.assign<int>(10,0x2D);
std::cout << str << '\n'; // "----------" str.assign(base.begin()+16,base.end()-12);
std::cout << str << '\n'; // "fox jumps over" return 0;
}

  

C++ -----> string:substr.

string substr (size_t pos = 0, size_t len = npos) const;
Generate substring

Returns a newly constructed string object with its value initialized to a copy of a substring of this object.

The substring is the portion of the object that starts at character position pos and spans len characters (or until the end of the string, whichever comes first).

Parameters

pos
Position of the first character to be copied as a substring.
If this is equal to the string length, the function returns an empty string.
If this is greater than the string length, it throws out_of_range.
Note: The first character is denoted by a value of 0 (not 1).
len
Number of characters to include in the substring (if the string is shorter, as many characters as possible are used).
A value of string::npos indicates all characters until the end of the string.

size_t is an unsigned integral type (the same as member type string::size_type).

609. Find Duplicate File in System的更多相关文章

  1. LC 609. Find Duplicate File in System

    Given a list of directory info including directory path, and all the files with contents in this dir ...

  2. 【leetcode】609. Find Duplicate File in System

    题目如下: Given a list of directory info including directory path, and all the files with contents in th ...

  3. 【LeetCode】609. Find Duplicate File in System 解题报告(Python & C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.c ...

  4. [LeetCode] Find Duplicate File in System 在系统中寻找重复文件

    Given a list of directory info including directory path, and all the files with contents in this dir ...

  5. [Swift]LeetCode609. 在系统中查找重复文件 | Find Duplicate File in System

    Given a list of directory info including directory path, and all the files with contents in this dir ...

  6. LeetCode Find Duplicate File in System

    原题链接在这里:https://leetcode.com/problems/find-duplicate-file-in-system/description/ 题目: Given a list of ...

  7. [leetcode-609-Find Duplicate File in System]

    https://discuss.leetcode.com/topic/91430/c-clean-solution-answers-to-follow-upGiven a list of direct ...

  8. Find Duplicate File in System

    Given a list of directory info including directory path, and all the files with contents in this dir ...

  9. HDU 3269 P2P File Sharing System(模拟)(2009 Asia Ningbo Regional Contest)

    Problem Description Peer-to-peer(P2P) computing technology has been widely used on the Internet to e ...

随机推荐

  1. ReboletricSample工程搭建

    受到  Just Say No to More End-to-End Tests 文章链接:http://googletesting.blogspot.tw/2015/04/just-say-no-t ...

  2. EasyDarwin开源流媒体项目

    本文转自EasyDarwin CSDN官方博客:http://blog.csdn.net/easydarwin EasyDarwin是由国内开源流媒体团队维护和迭代的一整套开源流媒体视频平台框架,从2 ...

  3. NOTE:rfc5766-turn-server

    NOTE:This project is active in Google code: http://code.google.com/p/rfc5766-turn-server/ 启动方法:./tur ...

  4. 九度OJ 1122:吃糖果 (递归)

    时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:1522 解决:1200 题目描述: 名名的妈妈从外地出差回来,带了一盒好吃又精美的巧克力给名名(盒内共有 N 块巧克力,20 > N ...

  5. java web service

    1.编写服务代码 服务代码提供了两个函数,分别为sayHello和sayHelloToPerson,源代码如下: /* * File name: HelloService.java * * Versi ...

  6. 周期性计划(一个cron守护进程):

    周期性计划(一个cron守护进程): root@ubuntu:/etc# ps -ef | grep cron root 903 1 0 16:25 ? 00:00:00 /usr/sbin/cron ...

  7. ARM编译器中预定义的宏

    arm系列目前支持三大主流的工具链,realview的armcc,iar ewarm的iccarm,gnu的gcc,编译器在编译的时候会预定义一些宏,这些宏在工程中起到不可或缺的作用. 例如 /* d ...

  8. 自动化测试框架selenium+java+TestNG——TestNG注解、执行、测试结果和测试报告

    TestNG是java的一个测试框架,相比较于junit,功能更强大和完善,我是直接学习和使用的TestNG就来谈下TestNG的一些特点吧. TestNG的特点 注解 TestNG使用Java和面向 ...

  9. (转)如何使用Java、Servlet创建二维码

    归功于智能手机,QR码逐渐成为主流,它们正变得越来越有用.从候车亭.产品包装.家装卖场.汽车到很多网站,都在自己的网页集成QR码,让人们快速找到它们.随着智能手机的用户量日益增长,二维码的使用正在呈指 ...

  10. UUID 和 GUID 的区别(转)

    UUID是一个由4个连字号(-)将32个字节长的字符串分隔后生成的字符串,总共36个字节长.比如:550e8400-e29b-41d4-a716-446655440000 http://gohands ...