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0388._Longest_Absolute_File_Path.md

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388. Longest Absolute File Path

难度: Medium

刷题内容

原题连接

内容描述

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 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.

解题方案

思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******

我们首先观察到每个文件夹或者是文件前面都会有一个'\n', 还有对应其层数个数的'\t'.

  • 所以首先根据'\n'分行，然后算出该文件/文件夹的层数depth
  • 如果是文件，我们需要更新maxlen
  • 如果是文件夹，我们需要更新该depth下的pathlen

程序变量解释

  • maxlen 代表目前最大子串的长度
  • pathlen 每一个depth下对应的path长度

特别需要注意的是，'\t'的长度是1

有的人仍然会有疑问，每次碰到文件夹都直接更新pathlen会不会导致本来长的反而变得短了，但是我们可以看到字符串的排版格式，每层path都是严格有自己的分级的， 因此不会出现这样的问题。 例如:

  • 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

其最大长度是32, "dir/subdir2/subsubdir2/file2.ext"

  • 如果变成"dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir20\n\t\tsubsubdir2\n\t\t\tfile2.ext"，
dir
    subdir1
        file1.ext
        subsubdir1
    subdir20
        subsubdir2
            file2.ext

最大长度就是33, "dir/subdir2/subsubdir20/file2.ext"

  • 如果变成 "dir\n\tsubdir1000000000000\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext"
dir
    subdir10000000000000
        file1.ext
        subsubdir1
    subdir20
        subsubdir2
            file2.ext

最大长度就是34,"dir/subdir10000000000000/file1.ext"

beats 99.66%

class Solution(object):
    def lengthLongestPath(self, input):
        """
        :type input: str
        :rtype: int
        """
        lap = 0
        depth_len = {0: 0}
        for line in input.splitlines():
            name = line.lstrip('\t')
            # 前面有几个'\t', depth就是几, 因为'\t'的长度为1
            depth = len(line) - len(name) 
            if '.' in name:
                lap = max(lap, depth_len[depth]+len(name))
            else:
                # 加1是为了加上一个path分隔符'/'的长度
                depth_len[depth+1] = depth_len[depth] + 1 + len(name) 
        return lap