LeetCode-Find Duplicate File in System

Description：
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?

题意：给定一个字符串数组，数组中的每个元素格式为“文件位置 文件1(内容) 文件2(内容)…文件n(内容)”；现在要求分组，将所有具有相同内容的路径划分为同一组（只有数量大于1才会划分组）；

解法：首先，需要找到相同内容的文件，这里可以利用哈希表来实现；我们定义哈希表map<key: 文件内容, value: 文件路径>；遍历字符串数组的所有元素path，以空格分割path所得到的第一个元素为文件的父路径dir，遍历除dir外的所有元素file（即文件），得到每个file的文件名fileName和文件内容content后，通过建立的哈希表有如下两个操作

• 如果表中已经存在key:content，则添加<key:content, value: oldValue + 文件路径>
• 如果表中不存在key:content，则添加<key:content, vlaue:文件路径>

我们最后得到的哈希表map中存储的就是以文件内容为键的，相同内容文件路径为键值的这样的一个表；最后只需要遍历这个表，取出所有分组数大于1的那些分组即可；

Java

class Solution {
    public List<List<String>> findDuplicate(String[] paths) {
        Map<String, List<String>> map = new HashMap<>();
        List<List<String>> res = new ArrayList<>();
        for (String path: paths) {
            String dir = path.substring(0, path.indexOf(' '));
            for (String file: path.split(" ")) {
                if (file.equals(dir)) {
                    continue;
                }
                String fileName = file.substring(0, file.indexOf('('));
                String content = file.substring(file.indexOf('(') + 1, file.indexOf(')'));
                if (map.containsKey(content)) {
                    List<String> temp = map.get(content);
                    temp.add(dir + "/" + fileName);
                    map.put(content, temp);
                } else {
                    List<String> temp = new ArrayList<>();
                    temp.add(dir + "/" + fileName);
                    map.put(content, temp);
                }
            }
        }
        for (String key: map.keySet()) {
            if (map.get(key).size() > 1) {
                res.add(map.get(key));
            }
        }

        return res;
    }
}