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LeetCode解题报告（379) -- 609. Find Duplicate File in System

Posted on Edited on Valine:

Problem

Given a list paths of directory info, including the directory path, and all the files with contents in this directory, return all the duplicate files in the file system in terms of their paths. You may return the answer in any order.

A group of duplicate files consists of at least two files that have 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 the 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 groups 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:

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Input: paths = ["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"]]

Example 2:

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

Constraints:

  • 1 <= paths.length <= 2 * 104
  • 1 <= paths[i].length <= 3000
  • 1 <= sum(paths[i].length) <= 5 * 105
  • paths[i] consist of English letters, digits, '/', '.', '(', ')', and ' '.
  • 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. A single blank space separates the directory path and file info.

Follow up:

  • 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?

Analysis

  这道题目是字符串处理的题目，题目看上去比较复杂，但实际上就是对字符串的切割。题目给出了一个字符串数组，每一个字符串都是包含三部分：文件路径，后面是若干个文件名+文件内容。最后题目要求返回的是，按照文件内容分类，把相同文件内容的路径合到一个数组。

  首先是利用空格，把文件路径截取出来，然后利用(把文件名拿到，最后再拿到文件内容。这里用一个map存储，利用文件内容作为key，value是路径，然后把文件路径加上文件名存进去。

Solution

  C++处理字符串时可以多用streamstring来切割，同时还可以使用find_last_of函数。

Code

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class Solution {
public:
    vector<vector<string>> findDuplicate(vector<string>& paths) {
        vector<vector<string>> res;
        unordered_map<string, vector<string>> m;
        for (string path : paths) {
            istringstream is(path);
            string pre = "", t = "";
            is >> pre;
            while (is >> t) {
                int idx = t.find_last_of('(');
                string dir = pre + "/" + t.substr(0, idx);
                string content = t.substr(idx + 1, t.size() - idx - 2);
                m[content].push_back(dir);
            }
        }
        for (auto a : m) {
            if (a.second.size() > 1)res.push_back(a.second);
        }
        return res;
    }
};

Summary

  这道题目理解起来有点复杂，但是整体的思路和方法都是很简单的，就是字符串的切割。这道题目的分享到这里，感谢你的支持！

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