[LeetCode 388] Longest Absolute File Path

最新推荐文章于 2020-05-31 11:50:34 发布

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本文介绍了一个算法问题，即在抽象的文件系统字符串中找到最长绝对路径的文件。通过使用vector数组记录不同深度的目录长度，文章详细解释了如何高效地解决此问题，达到O(n)的时间复杂度。

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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. subdir1contains 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.

分析

这一题题目挺唬人的，但是如果比较细心的话，应该还是能分析能够写出来的。

这题的关键在于我们要记录之前的目录的长度，我们使用vector 数组记录每一个depth的目录的长度，第i位表示当前深度为i的目录的长度。每当遇到\n时我们就知道新换了一行，判断之前记录的是文件还是目录，如果是目录，就将目录的长度记录到数组的depth位，如果是文件，就统计总长度，之后将depth清空，每次遇到\t时就将depth ++。当遇到'.'时，就记录这是一个文件。

当然也可以使用stack记录depth的目录长度，但是相比于数组，stack的实现的复杂度要高，因为depth减小时，需要将大于depth的项全部弹出。

Code

class Solution {
public:
    int lengthLongestPath(string input) {
        int res = 0;
          int len = input.size();
          if (len == 0)return 0;
      
          vector<int> dirLen;
          bool isFile = false;
          int depth = 0;
          int wordLen = 0;
          input += '\n';
          len = input.size();
      
          for (int i = 0; i < len; i ++) 
          {   
              if (input[i] == 10) 
              {   
                  if (depth >= dirLen.size())
                  {   
                      dirLen.push_back(wordLen);
                  }   
                  else
                  {   
                      dirLen[depth] = wordLen;
                  }   
                  if (isFile)
                  {   
                      int sum = 0;
                      for (int j=0; j <= depth; j ++) 
                      {   
                          sum += dirLen[j];
                      }   
                      res = max(res, sum + depth);
                  }   
                  cout << i << '\t' << depth << '\t' << wordLen << endl;
                  wordLen = 0;
                  depth = 0;
                  isFile = false;
              }   
              else if (input[i] == 9)
              {   
                  depth += 1;
              }   
              else if (input[i] == '.')
              {   
                  isFile = true;
                  wordLen ++; 
              }   
              else
              {   
                  wordLen ++; 
              }   
          }   
  
          return res;

    }
};

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