388. Longest Absolute File Path

问题描述

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-directoriessubdir1 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.

题目链接：

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思路分析

给一个string，记录了一个文件系统的结构，用\n表示下一行，\t表示tab，求这个string代表的文件系统中最长的路径。注意‘\’也算做一部分。

用一个dirs向量保持此时遍历时的文件路径的长度，level记录应该存放的位置：当出现\n时，表示换下一行，level，len，isfile都要重新记；当出现\t时，表示层级++；当出现 . 时，则是一个文件，可以进行长度计算了。记录的时0到level的文件名，还有文件，所以要accumulate到level+1，accumulate需要初始值为0，然后再加上分隔每个文件的\，一共有level个，就是路径长度了。

代码

class Solution {
public:
    int lengthLongestPath(string input) {
        vector<int> dirs(256, 0);
        int res = 0;
        input.push_back('\n');

        for (int i = 0, level = 0, len = 0, isfile = 0; i < input.size(); i++){
            switch (input[i]){
                case '\n': level = 0; len = 0; isfile = 0; break;
                case '\t': level++; break;
                case '.' : isfile = 1;
                default :
                    len++;
                    dirs[level] = len;
                    if (isfile)
                        res = max(res, accumulate(dirs.begin(), dirs.begin() + level + 1, 0) + level);
            }
        }
        return res;
    }
};

时间复杂度：$O(n)$
空间复杂度：$O(n)$

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反思

是一道挺费我脑子的题，学了很多，accumulate的用法；vector的初始化dirs（256个，初始值为0）；string中的转义字符。