609. Find Duplicate File in System
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.txt
, f2.txt
... fn.txt
with content f1_content
, f2_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:
- No order is required for the final output.
- 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].
- The number of files given is in the range of [1,20000].
- You may assume no files or directories share the same name in the same directory.
- 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:
- Imagine you are given a real file system, how will you search files? DFS or BFS?
- If the file content is very large (GB level), how will you modify your solution?
- If you can only read the file by 1kb each time, how will you modify your solution?
- 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?
- How to make sure the duplicated files you find are not false positive?
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> |
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;
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
- 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 ...
- 【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 ...
- 【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 ...
- [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 ...
- [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 ...
- 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 ...
- [leetcode-609-Find Duplicate File in System]
https://discuss.leetcode.com/topic/91430/c-clean-solution-answers-to-follow-upGiven a list of direct ...
- Find Duplicate File in System
Given a list of directory info including directory path, and all the files with contents in this dir ...
- 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
- Android UI经常使用实例 怎样实现欢迎界面(Splash Screen)
在Android平台下.下载一个应用后,首次打开映入眼帘的便是Splash Screen,暂且不说Android的设计原则提不提倡这样的Splash Screen.先来看看一般使用Splash Scr ...
- Java多态特性:重载和覆写的比較
Java重载: 在同一个类中 方法具有同样的名字,同样或不同的返回值,但參数不同的多个方法(參数个数或參数类型) public class MethoDemo{ public static void ...
- OSEck中odo_vect2pcb的作用
在基于OSEck RTOS的TI DSP中,中断能够作为一个进程存在,在OSEck系统中,进程分为两类:优先级进程,中断进程. 当可屏蔽中断(INT4~15)发生后,就会运行相应的中断vector,在 ...
- mysql-connector-java与mysql版本的对应
记录下mysql-connector-java与mysql版本的对应关系,已方便以后参考,这是最新版本对应, 时间:2017年5月23日 官网文档地址: https://dev.mysql.com/d ...
- java object monitor
1 什么是java object monitor 每个java对象头中都有锁状态位标记.java中在使用synchronize同步的时候,肯定是涉及到某个对象的锁.因此,在考虑同步的时候,首先要想到是 ...
- mongodb学习之:聚合
@font-face { font-family: "Times New Roman"; }@font-face { font-family: "宋体"; }p ...
- DTLS学习笔记 -- RFC 4347- 6347
想学习一下dtls,是因为想以后没有公司免费VPN可用的时候,我能买一个主机,自己建一个VPN. 1.介绍 Web, email大多用TLS协议来做安全的网络传输,它们必须跑在可靠的TCP传输通道里. ...
- [haoi2014]贴海报
Bytetown城市要进行市长竞选,所有的选民可以畅所欲言地对竞选市长的候选人发表言论.为了统一管理,城市委员会为选民准备了一个张贴海报的electoral墙.张贴规则如下:1.electoral墙是 ...
- 转载——Android permission 访问权限大全
程序执行需要读取到安全敏感项必需在androidmanifest.xml中声明相关权限请求, 完整列表如下: Android.permission.ACCESS_CHECKIN_PROPERTIES ...
- XML中CDATA和#PCDATA的区别
在XML文档中, 能看到“CDATA"的地方有三处: 1)在DTD中,指定标签中某个属性的类型为字符型时,使用CDATA.因为XML解析器会去分析这段字符内容,因而里面如果需要使用>, ...