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(C++)
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?Solution 1(C++)
class Solution {
public:
vector<vector<string>> findDuplicate(vector<string>& paths) {
unordered_map<string, vector<string>> files;
vector<vector<string>> result;
for (auto path : paths) {
stringstream ss(path);
string root;
string s;
getline(ss, root, ' ');
while (getline(ss, s, ' ')) {
string fileName = root + '/' + s.substr(0, s.find('('));
string fileContent = s.substr(s.find('(') + 1, s.find(')') - s.find('(') - 1);
files[fileContent].push_back(fileName);
}
}
for (auto file : files) {
if (file.second.size() > 1)
result.push_back(file.second);
}
return result;
}
};算法分析
算法层面不难，主要是函数的使用。
程序分析
这里涉及到了STL的IO库中的< sstream > 库，主要有三种类：istringstream、ostringstream和stringstream，分别用来进行流的输入、输出和输入输出操作。具体的详细试用我会单独整理成一篇博客。这里就主要说明这个程序中stringstream的作用。
这里使用stringstream主要是将字符串分割开来。
stringstream sstrm(s);sstrm是一个sstream对象，保存string s的一个拷贝，此构造函数是explicit的。所以这里是将string path构造为一个sstream ss。然后利用getline()函数：
#include <string>
istream& getline ( istream &is , string &str , char delim );其中
- istream &is 表示一个输入流，譬如cin，这里使用stringstream ss；
- string&str表示把从输入流读入的字符串存放在这个字符串中，比如这里构造的string s；
- char delim表示遇到这个字符停止读入，在不设置的情况下系统默认该字符为’\n’，也就是回车换行符（遇到回车停止读入）。这里设置为空格字符：’ ‘。
经过这个函数之后，root将会保存根目录路径。然后就是利用类似的方法获取后面的内容。多多注意getline()函数的方法。