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609. Find Duplicate File in System

时间:2017-10-18 16:09:50      阅读:201      评论:0      收藏:0      [点我收藏+]

标签:bcd   code   represent   ace   txt   min   ++   positive   ons   

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?

题目含义:这道题给了我们一堆字符串数组,每个字符串中包含了文件路径,文件名称和内容,让我们找到重复的文件,这里只要文件内容相同即可,不用管文件名是否相同,而且返回结果中要带上文件的路径

 1     public List<List<String>> findDuplicate(String[] paths) {
 2         List<List<String>> result = new ArrayList<List<String>>();
 3         int n = paths.length;
 4         if (n == 0) return result;
 5         
 6         Map<String, Set<String>> map = new HashMap<>();
 7         for (String path : paths) {
 8             String[] strs = path.split("\\s+");
 9             for (int i = 1; i < strs.length; i++) {
10                 int idx = strs[i].indexOf("(");
11                 String content = strs[i].substring(idx);
12                 String filename = strs[0] + "/" + strs[i].substring(0, idx);
13                 Set<String> filenames = map.getOrDefault(content, new HashSet<String>());
14                 filenames.add(filename);
15                 map.put(content, filenames);
16             }
17         }
18         
19         for (String key : map.keySet()) {
20             if (map.get(key).size() > 1) {
21                 result.add(new ArrayList<String>(map.get(key)));
22             }
23         }
24         
25         return result;     
26     }

 

609. Find Duplicate File in System

标签:bcd   code   represent   ace   txt   min   ++   positive   ons   

原文地址:http://www.cnblogs.com/wzj4858/p/7686984.html

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