问题：
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.
解决：
【题意】我们的目的是找到文件系统中文件的最长（字符数）绝对路径。例如，在上面的第二个例子中，最长的绝对路径是“dir / subdir2 / subsubdir2 / file2.ext”，其长度是32（字符的长度，不包括双引号）。
【注】1. 文件的名称至少包含一个 . 和一个扩展名。
2. 目录或子目录的名称将不包含.。 3. 所需的时间复杂度：O（n）其中n是输入字符串的大小。如果有另一个路径aaaaaaaaaaaaaaaaaaaaa / sth.png， a / aa / aaa / file1.txt不是最长的文件路径。
① 利用\n来分割文件夹和文件名，利用\t来统计此时文件夹或文件对应的层数，然后利用哈希表来建立深度和当前深度的绝对路径长度之间的映射。
找到'\n'时判断当前深度是文件还是文件夹，找到‘\t’深度自增一(每次遍历到'\n',要level=0,因为'\t'就是相对应深度)。
class Solution { //4ms
public int lengthLongestPath(String input) { String[] paths = input.split("\n");//该深度结束 int[] tmp = new int[paths.length + 1]; int maxlen = 0; for (String str : paths){ int level = str.lastIndexOf("\t") + 1;//深度自增1 tmp[level + 1] = tmp[level] + str.length() - level + 1;//计算路径长度 if (str.contains(".")){ maxlen = Math.max(maxlen,tmp[level + 1] - 1); } } return maxlen; } }