[Leetcode] 388. Longest Absolute File Path 解题报告

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本文介绍了一种使用栈来解决寻找文件系统中具有最长绝对路径文件的方法。通过解析输入字符串并跟踪目录层级，该算法能够在O(n)的时间复杂度内找到最长路径。

题目：

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.

思路：

借助一个栈来保存当前层的路径，层数可以利用“\t”的个数来确定，如果当前行的层数大于栈顶元素并且非文件，就可以让其入栈；如果当前行是文件就可以比较大小。如果当前行是目录但是深度小于等于栈顶元素，需要将栈顶元素出栈，直到为空或者小于当前行的深度。

代码：

class Solution {
public:
    int lengthLongestPath(string input) {
        input += "\n";
        int len = input.size(), ans = 0;
        int space_count = 0;
        int cur_path_len = 0;
        bool dot_flag = false;
        stack<pair<string, int>> st;
        string cur_dir, extension;
        for (int i = 0; i < len; ++i) {
            if (input[i] != '\n') {
                if (input[i] == '\t') {
                    space_count++;
                }
                else {
                    cur_dir += input[i];
                }
                if (dot_flag) {
                    extension += input[i];
                }
                if (input[i] == '.') {
                    dot_flag = true;
                }
            }
            else {      // we need to come to another line
                while (!st.empty() && space_count <= st.top().second) {         // pop up the extra folders
                    cur_path_len -= st.top().first.size();
                    st.pop();
                }
                if (dot_flag) {
                    ans = max(cur_path_len + (int)cur_dir.size() + 1, ans);
                }
                else {
                    st.push(make_pair("/" + cur_dir, space_count));
                    cur_path_len += cur_dir.size() + 1;
                }
                extension = "", dot_flag = false, space_count = 0, cur_dir = "";
            }
        }
        return ans == 0 ? 0 : ans - 1;
    }
};