Longest Absolute File Path问题及解法

问题描述：

Suppose we abstract our file system by a string in the following manner:

The string 

"dir\n\tsubdir1\n\tsubdir2\n\t\tfile.ext"

 represents:

dir
subdir1
subdir2
file.ext

The directory 

dir

 contains an empty sub-directory 

subdir1

 and
a sub-directory 

subdir2

 containing a file 

file.ext

.

The string 

"dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext"

represents:

dir
subdir1
file1.ext
subsubdir1
subdir2
subsubdir2
file2.ext

The directory 

dir

 contains two sub-directories 

subdir1

 and 

subdir2

. 

subdir1

 contains
a file 

file1.ext

 and an empty second-level sub-directory 

subsubdir1

. 

subdir2

 contains
a second-level sub-directory 

subsubdir2

 containing a file 

file2.ext

.

We are interested in finding the longest (number of characters) absolute path to a file within our file system. For example, in the second example above, the longest absolute path is 

"dir/subdir2/subsubdir2/file2.ext"

,
and its length
ae25
is 

32

 (not including the double quotes).

Given a string representing the file system in the above format, return the length of the longest absolute path to file in the abstracted file system. If there is no file in the system, return 

0

.

Note:

The name of a file contains at least a 

.

 and an extension.
The name of a directory or sub-directory will not contain a 

.

.

Time complexity required: 

O(n)

 where 

n

 is
the size of the input string.

Notice that 

a/aa/aaa/file1.txt

 is not the longest file path, if there is another path 

aaaaaaaaaaaaaaaaaaaaa/sth.png

.

问题分析：

我们可以利用'\t'的个数来决定文件或是文件夹在那一层，当读到的是文件时，我们可以根据保存的路径得到路径长度。

过程相见代码：

class Solution {
public:
int lengthLongestPath(string input) {
vector<string> path;
int res = 0;
int depth = 0; // 代表t的个数
bool flag = false; // 检验是文件还是文件夹，默认是文件夹
int count = 0;// 统计一个文件或一个文件夹的长度

for (int i = 0; i <= input.length(); i++)
{
if (i == input.length() || input[i] == '\n')
{
string str = input.substr(i - count, count);
if (path.size() == depth) path.emplace_back(str);

else path[depth] = str;

if (flag)
{
int len = depth;
for (int j = 0; j <= depth; j++)
{
len += path[j].length();
}
if (res < len) res = len;
flag = false;
}
count = 0;
depth = 0;
continue;
}
if (input[i] == '\t')
{

while (input[i] == '\t')
{
i++;
depth++;
}
i--;
continue;
}
if (input[i] == '.') flag = true;
count++;
}

return res;
}
};