388. Longest Absolute File Path (Medium)

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.

Solution:

这道题给了我们一个字符串，里面包含\n和\t这种表示回车和空格的特殊字符，让我们找到某一个最长的绝对文件路径，要注意的是，最长绝对文件路径不一定是要最深的路径，我们可以用哈希表来建立深度和当前深度的绝对路径长度之间的映射，那么当前深度下的文件的绝对路径就是文件名长度加上哈希表中当前深度对应的长度，我们的思路是遍历整个字符串，遇到\n或者\t就停下来，然后我们判断，如果遇到的是回车，我们把这段文件名提取出来，如果里面包含'.'，说明是文件，我们更新res长度，如果不包含点，说明是文件夹，我们深度level自增1，然后建立当前深度和总长度之间的映射，然后我们将深度level重置为0。之前如果遇到的是空格\t，那么我们深度加一，通过累加\t的个数，我们可以得知当前文件或文件夹的深度，然后做对应的处理，参见代码如下：

int lengthLongestPath(string input) {
    int res = 0, n = input.size(), level = 0;
    unordered_map<int, int> m{{0,0}};
    for (int i = 0; i < n; ++i) {
        int start = i;
        // get file or dir name
        while (i < n && input[i] != '\n' && input[i] != '\t') ++i;

        // reach line end, set level to 0
        if (i >= n || input[i] == '\n') {
            string t = input.substr(start, i-start);
            if (t.find('.') != string::npos) {
                // is dir
                res = max(res, m[level]+(int)t.size());
            } else {
                ++level;
                m[level] = m[level-1]+(int)t.size()+1;
            }
            level = 0;
        } else {
            // '\t' increase depth
            ++level;
        }
    }
    return res;
}

use stringstream:

int lengthLongestPath(string input) {
    int res = 0;
    istringstream ss(input);
    unordered_map<int, int> m{{0,0}};
    string line;
    while (getline(ss, line)) {
        int level = line.find_last_of('\t')+1;
        int len = line.substr(level).size();
        if (line.find('.') != string::npos) {
            res = max(res, m[level]+len);
        } else {
            m[level+1] = m[level]+len+1;
        }
    }
    return res;
}

use vector:

int lengthLongestPath(string input) {
    int res = 0;
    istringstream ss(input);
    vector<int> m(1,0);
    string line;
    while (getline(ss, line)) {
        int level = line.find_last_of('\t')+1;
        int len = line.substr(level).size();
        if (line.find('.') != string::npos) {
            res = max(res, m[level]+len);
        } else {
            if (m.size() == level+1) m.push_back(m[level]+len+1);
            else m[level+1] = m[level]+len+1;
        }
    }
    return res;
}