题目地址：https://leetcode.com/problems/find-duplicate-file-in-system/description/
题目描述
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.
Agroup of duplicate files consists of at least two files that have exactly the same content.
Asingle 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)"
Itmeans 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.
Theoutput 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、 Noorderisrequiredforthefinaloutput.；
2、 Youmayassumethedirectoryname,filenameandfilecontentonlyhaslettersanddigits,andthelengthoffilecontentisintherangeof[1,50].；
3、 Thenumberoffilesgivenisintherangeof[1,20000].；
4、 Youmayassumenofilesordirectoriessharethesamenameinthesamedirectory.；
5、 Youmayassumeeachgivendirectoryinforepresentsauniquedirectory.Directorypathandfileinfoareseparatedbyasingleblankspace.；
Follow-up beyond contest:
1、 Imagineyouaregivenarealfilesystem,howwillyousearchfiles?DFSorBFS?；
2、 Ifthefilecontentisverylarge(GBlevel),howwillyoumodifyyoursolution?；
3、 Ifyoucanonlyreadthefileby1kbeachtime,howwillyoumodifyyoursolution?；
4、 Whatisthetimecomplexityofyourmodifiedsolution?Whatisthemosttime-consumingpartandmemoryconsumingpartofit?Howtooptimize?；
5、 Howtomakesuretheduplicatedfilesyoufindarenotfalsepositive?；
题目大意
把不同文件夹中所有文件内容相同的文件放到一起。
解题方法
这个题很简单，只需要使用字典进行内容==>目录的对应保存即可。因为要得到内容相同的目录的列表，所以把内容作为键，把目录列表作为值。最后的结果要目录列表内容长度>1才行。
Python代码：
class Solution(object):
def findDuplicate(self, paths):
"""
:type paths: List[str]
:rtype: List[List[str]]
"""
filemap = collections.defaultdict(list)
for path in paths:
roads = path.split()
directory, files = roads[0], roads[1:]
for file in files:
file_s = file.split('(')
name, content = file_s[0], file_s[1][:-1]
full = directory + '/' + name
filemap[content].append(full)
return [full for full in filemap.values() if len(full) > 1]
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这个题的C++解法让我学习到了istringstream的用法，istringstream是一个比较有用的c++的输入输出控制类。
C++引入了ostringstream、istringstream、stringstream这三个类，要使用他们创建对象就必须包含<sstream>这个头文件。 istringstream类用于执行C++风格的串流的输入操作。 ostringstream类用于执行C风格的串流的输出操作。 strstream类同时可以支持C风格的串流的输入输出操作。
和常见的iostream有点类似，可以对应理解。
C++代码如下：
class Solution {
public:
vector<vector<string>> findDuplicate(vector<string>& paths) {
unordered_map<string, vector<string>> m;
vector<vector<string>> res;
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;
}
};
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参考资料：
http://www.cnblogs.com/grandyang/p/7007974.html https://blog.csdn.net/longzaitianya1989/article/details/52909786
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