388. Longest Absolute File 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
.
Solution
(1) Java
class Solution {
public int lengthLongestPath(String input) {
if (input == null || input.length() == 0) {
return 0;
}
String[] paths = input.split("\n");
Deque<Integer> stack = new LinkedList<>();
int rst = 0;
int idx = 0;
for (String path : paths) {
if (path.indexOf('\t') == -1) {
if (path.indexOf('.') != -1) {
rst = Math.max(rst, path.length());
} else {
while (stack.size() > 0) {
stack.pop();
}
stack.push(path.length());
}
} else {
int level = 0;
while (path.indexOf('\t') != -1) {
level++;
path = path.substring(1);
}
while(stack.size() > level) {
stack.pop();
}
if (path.indexOf('.') != -1) {
rst = Math.max(rst, stack.peek()+path.length()+1);
} else {
stack.push(path.length()+stack.peek()+1);
}
}
}
return rst;
}
}
Same algorithm, but more concise code
public int lengthLongestPath(String input) {
Deque<Integer> stack = new ArrayDeque<>();
stack.push(0); // "dummy" length
int maxLen = 0;
for(String s:input.split("\n")){
int lev = s.lastIndexOf("\t")+1; // number of "\t"
while(lev+1<stack.size()) stack.pop(); // find parent
int len = stack.peek()+s.length()-lev+1; // remove "/t", add"/"
stack.push(len);
// check if it is file
if(s.contains(".")) maxLen = Math.max(maxLen, len-1);
}
return maxLen;
}
(2) Python
class Solution:
def lengthLongestPath(self, input):
"""
:type input: str
:rtype: int
"""
if not input:
return 0
paths = input.split("\n")
stack = []
rst = 0
for path in paths:
level = path.rfind('\t') + 1
while len(stack) > level:
stack.pop()
length = len(path)-level
if stack:
length += stack[-1]+1
stack.append(length)
if path.find('.') != -1:
rst = max(rst, length)
return rst
(3) Scala