# 388. Longest-Absolute-File-Path

difficulty: Medium

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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`.

## Method One

```java
class Solution {
    public int lengthLongestPath(String input) {
        // Note: \n \t 叫 escaped characters，长度是1。

        // 这个题有 tree-structure 自然想到用 tree的DFS遍历
        // 但是它只是一个String，写法上要有区别

        // 注意到 每个 \n 都意味着一个新的元素，所以按 \n 分  

        // 注意，如果没有文件的话，长度是0.
        LinkedList<Integer> pathLength = new LinkedList<>();
        pathLength.push(0); // dummy length, and help to increase the size.
        int maxLength = 0;
        for(String path : input.split("\n")) {
            int level = path.lastIndexOf("\t") + 1; // number of \t no \t would return -1 + 1 = 0;
            // find correct father
            while( level + 1 < pathLength.size() ) {
                pathLength.pop();
            }
​
            int newLength = pathLength.peek() + path.length() - level + 1; // - \t + /

            if(path.contains(".")) {
                newLength -= 1;
                maxLength = Math.max(maxLength, newLength);
            }

            pathLength.push(newLength);

        }

        return maxLength;

    }

}
​
​
​
```
