388. Longest Absolute File Path

原题链接： https://leetcode.com/problems/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.

 

几个基础知识，关于string的

1. '\n','\t'这种都算作一个char，长度都是1

2. str = "\t\t\tabc", str.lastIndexOf('\t) = 2. 从0记起

 

这道题的思路是用一个stack记录父节点，当遍历到一个level是n的文件的时候，stack里面应该有n-1个文件层次的，如果多了就弹出。

但是需要开始推一个dummy的头进去，就是假设有一个-1层，它的长度是0

    public int lengthLongestPath(String input) {
        if(input == null || input.length() == 0) {
            return 0;
        }   
        Stack<Integer> stack = new Stack<>();
        stack.push(0);
        int maxLen = 0;
        for(String str: input.split("\n")) {
            int level = str.lastIndexOf('\t') + 1;// 也就是\t的个数
            while(stack.size() != level + 1) {
                stack.pop(); 
            }
            int len = stack.peek() + str.length() - level + 1; //把\t去掉，加一个/
            stack.push(len);
            if(str.indexOf('.') != -1) {
                maxLen = Math.max(maxLen, len - 1);
            }
        }
        return maxLen;
    }

时间复杂度是O(n)