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-directorysubsubdir1
. 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
.
這道題給了我們一個字符串,里面包含 \n 和 \t 這種表示回車和空格的特殊字符,讓我們找到某一個最長的絕對文件路徑,要注意的是,最長絕對文件路徑不一定是要最深的路徑,我們可以用 HashMap 來建立深度和當前深度的絕對路徑長度之間的映射,那么當前深度下的文件的絕對路徑就是文件名長度加上 HashMap 中當前深度對應的長度,我們的思路是遍歷整個字符串,遇到 \n 或者 \t 就停下來,然后我們判斷,如果遇到的是回車,我們把這段文件名提取出來,如果里面包含 '.',說明是文件,我們更新 res 長度,如果不包含點,說明是文件夾,我們深度 level 自增1,然后建立當前深度和總長度之間的映射,然后我們將深度 level 重置為0。之前如果遇到的是空格 \t,那么我們深度加一,通過累加 \t 的個數,我們可以得知當前文件或文件夾的深度,然后做對應的處理,參見代碼如下:
C++ 解法一:
class Solution { public: int lengthLongestPath(string input) { int res = 0, n = input.size(), level = 0; unordered_map<int, int> m {{0, 0}}; for (int i = 0; i < n; ++i) { int start = i; while (i < n && input[i] != '\n' && input[i] != '\t') ++i; if (i >= n || input[i] == '\n') { string t = input.substr(start, i - start); if (t.find('.') != string::npos) { res = max(res, m[level] + (int)t.size()); } else { ++level; m[level] = m[level - 1] + (int)t.size() + 1; } level = 0; } else { ++level; } } return res; } };
下面這種方法用到了字符串流機制,通過 getline() 函數可以一行一行的獲取數據,實際上相當於根據回車符 \n 把每段分割開了,然后對於每一行,我們找最后一個空格符 \t 的位置,然后可以得到文件或文件夾的名字,然后我們判斷其是文件還是文件夾,如果是文件就更新 res,如果是文件夾就更新 HashMap 的映射,參見代碼如下:
C++ 解法二:
class Solution { public: int lengthLongestPath(string input) { int res = 0; istringstream ss(input); unordered_map<int, int> m{{0, 0}}; string line = ""; while (getline(ss, line)) { int level = line.find_last_of('\t') + 1; int len = line.substr(level).size(); if (line.find('.') != string::npos) { res = max(res, m[level] + len); } else { m[level + 1] = m[level] + len + 1; } } return res; } };
Java 解法二:
public class Solution { public int lengthLongestPath(String input) { int res = 0; Map<Integer, Integer> m = new HashMap<>(); m.put(0, 0); for (String s : input.split("\n")) { int level = s.lastIndexOf("\t") + 1; int len = s.substring(level).length(); if (s.contains(".")) { res = Math.max(res, m.get(level) + len); } else { m.put(level + 1, m.get(level) + len + 1); } } return res; } }
參考資料:
https://leetcode.com/problems/longest-absolute-file-path/
https://leetcode.com/problems/longest-absolute-file-path/discuss/86615/9-lines-4ms-Java-solution
https://leetcode.com/problems/longest-absolute-file-path/discuss/86821/c-on-solution-with-hashmap