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?

思路:content为key,文件路径为value,全部遍历存入map即可.没有用到什么树型表示文件系统那么高端的东西.重点在于字符串的分割,以及用stringstream配合getline()来高效切分字符串的技巧.

至于map,之前想用multimap,后来发现输出不符合题目要求,因为最终要输出的是一个双重vector,可重map反而麻烦.所以还是用复合map比较恰当.而且是用unordered_map,速度比map快些(不用排序).

vector<vector<string>> findDuplicate(vector<string>& paths) {
unordered_map<string, vector<string>> files;
vector<vector<string>> res;
for (auto path : paths) {
stringstream ss(path);  //转成字符串流,便于用getline()切割
string root;
getline(ss, root, ' ');  //以空格为分界,切割出路径存于root
string s;  //用于存放切割出的文件名
while (getline(ss, s, ' ')) {
string fileName = root + '/' + s.substr(0, s.find('('));  //切割出文件名
string fileContent = s.substr(s.find('(')+1, s.find(')')-s.find('(')-1);  //取出文件内容
files[fileContent].push_back(fileName);
}
}
for (auto file : files) {
if (file.second.size() > 1) {  //大于1表示有重复的
res.push_back(file.second);
}
}
return res;
}