A mistake to come up a solution is to forget to add a clash into the absolute path.
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 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
.1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 | public class Solution { public int lengthLongestPath(String input) { //sequencially going through all the paths and files, //not care about the empty dirs because only file contribute to file path length String[] dirs = input.split("\n");//parse out each directory Stack<Integer> lvls= new Stack<Integer>(); int maxLen = 0; for(String s:dirs){ int lev = s.lastIndexOf("\t"); if(lev==-1){//this must be direct children in root dir lvls.clear(); lvls.push(s.length()+1);//every push is a path, and +1 for a slash } else{ //for subdirs or files //want to retreat back to parent level while(lvls.size()>lev+1)lvls.pop(); //lvl is -1, 0,1 etc, with 0 means first sublevel. because when its size is greater than lvl, //it will go further to reduce one, so lev+1 make sure it stops at parent level/lev after the further reduction lvls.push(lvls.peek()+s.length()-(lev+1)+1);//lev+1 is number if "\t" } if(s.contains(".")) maxLen = Math.max(maxLen, lvls.peek()-1);//file is not ended with a slash } return maxLen; } } |
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