0609. Find Duplicate File in System

# 609. Find Duplicate File in System#

## 题目 #

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:

``````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:

``````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.

• 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?

## 解题思路 #

• 这一题算简单题，考察的是字符串基本操作与 map 的使用。首先通过字符串操作获取目录路径、文件名和文件内容。再使用 map 来寻找重复文件，key 是文件内容，value 是存储路径和文件名的列表。遍历每一个文件，并把它加入 map 中。最后遍历 map，如果一个键对应的值列表的长度大于 1，说明找到了重复文件，可以把这个列表加入到最终答案中。
1. 假设您有一个真正的文件系统，您将如何搜索文件？广度搜索还是宽度搜索？
2. 如果文件内容非常大（GB级别），您将如何修改您的解决方案？
3. 如果每次只能读取 1 kb 的文件，您将如何修改解决方案？
4. 修改后的解决方案的时间复杂度是多少？其中最耗时的部分和消耗内存的部分是什么？如何优化？
5. 如何确保您发现的重复文件不是误报？

## 代码 #

``````package leetcode

import "strings"

func findDuplicate(paths []string) [][]string {
cache := make(map[string][]string)
for _, path := range paths {
parts := strings.Split(path, " ")
dir := parts[0]
for i := 1; i < len(parts); i++ {
bracketPosition := strings.IndexByte(parts[i], '(')
content := parts[i][bracketPosition+1 : len(parts[i])-1]
cache[content] = append(cache[content], dir+"/"+parts[i][:bracketPosition])
}
}
res := make([][]string, 0, len(cache))
for _, group := range cache {
if len(group) >= 2 {
res = append(res, group)
}
}
return res
}
``````

Apr 8, 2023