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.

TIPS:

stack 和 array的解法差不多，下面是理解：都是O(n)因为所有的东西只会进一次出一次。

I'm assuming that "\t"will appear only before the directory or file name.

For example, if s = "\t\t\tdirname", then s.lastIndexOf("\t") will be 2, the number of "\t" will be 3. If a diretory name does not contain"\t", then s.lastIndexOf("\t") will be -1, the number of "\t" will be 0.

However, if "\t" is allowed within the directory or file name, then this way does not work.

用\t的个数来代表层数，\n来区分文件。

Code:

Array:(更快）

public class Solution {
    public int lengthLongestPath(String input) {
        String[] files = input.split("\n");
        int[] stack = new int[files.length + 1];
        int maxLength = 0;
        stack[0] = 0;
        for (String s : files) {
            int level = s.lastIndexOf("\t") + 1;
            int curLength = stack[level] + s.length() - level + 1;
            stack[level + 1] = curLength;
            if (s.contains(".")) {
                maxLength = Math.max(maxLength, curLength - 1);
            }
        }
        return maxLength;
    }
}

stack：

  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;
      }