388. 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. subdir2contains 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.

首先想到的是用stack，存外层的长度，对输入的字符串根据“\n”分割之后，含有几个“\t”就是第几层。遍历下去，如果层数比stack的size小，说明上面有跟这层没关系的，全pop掉，然后更新现在的len，如果当前字符串包含‘.’，更新max。代码如下：

public class Solution {
    public int lengthLongestPath(String input) {
        Stack<Integer> stack = new Stack<>();
        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;
    }
}

更快的方法可以用空间换时间，用一个lens数组保存每一层的长度，每次都会对应层的长度，上面如果有空的文件夹，在下面的更新中会被更新掉，保证每次保留的都是自己的父路径。代码如下：

public class Solution {
    public int lengthLongestPath(String input) {
        String[] strs = input.split("\n");
        int[] lens = new int[strs.length + 1];
        int currLen = 0, max = 0;
        for (String str: strs) {
            int lev = str.lastIndexOf('\t') + 1;
            currLen = lens[lev + 1] = lens[lev] + str.length() - lev + 1;
            if (str.contains(".")) {
                max = Math.max(max, currLen - 1);
            }
        }
        return max;
    }
}