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:

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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:

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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.

思路

利用栈来处理层次的问题。就像函数的调用一样。所以一开始下笔码代码的时候一定要想清楚怎么做，拿什么样的数据结构来做，不然会一团乱麻。

利用stringbuffer存储临时的string。这样可以降低时间复杂度。

需要判断这个string是不是文件，所以需要判断’.’是否在string当中。

需要判断‘\t’在不在这个string当中，有几个’\t’。

需要利用栈来存储string的长度数据。在栈当中，我们利用数据为负数，来表示这个字符串是文件。但是利用这样的方法存储‘是否为文件’的状态，一定要注意求取绝对值，不然就会影响答案的准确性。

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public class Solution {
    public int lengthLongestPath(String input) {
        StringBuffer tmp = new StringBuffer();
        // List<String> result = new ArrayList<String>();
        // List<String> result_tmp = new ArrayList<String>();
        // int result_tmp_length = 0;
        Stack<Integer> result = new Stack<Integer>();
        boolean isFile = false;
        int result_length = 0;
        int layer_count = 0;
        int final_length = 0;
        for (int i = 0; i < input.length(); i++){
            char cur = input.charAt(i);
            if(cur == '.') isFile = true;
            if(cur == '\t') layer_count++;
            if(cur != '\n'){
                tmp.append(cur);
            }
            if(cur == '\n' || i == input.length() - 1){
                //Calculate the temporary word length.
                String cur_tmp = tmp.toString();
                tmp.setLength(0);
                int cur_length = cur_tmp.length() - layer_count + 1;
                if(isFile) cur_length = -cur_length;
                isFile = false;
                //Reset the status of result Stack
                if(result.size() > layer_count){
                    int nums = result.size() - layer_count;
                    for(int j = 0; j < nums ; j++){
                        if(result.peek() < 0) result_length += result.pop();
                        else result_length -= result.pop();
                    }
                }
                result.push(cur_length);
                if(cur_length < 0){
                    result_length += (-cur_length);
                    if(result_length > final_length) final_length = result_length;
                }
                else{
                    result_length += cur_length;
                }
                //Initialize other variables
                layer_count = 0;
            }
        }
        if(final_length == 0) return 0;
        else return final_length - 1;
    }
}