[LeetCode] 388. Longest Absolute File Path 最长的绝对文件路径

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-directorysubsubdir1. 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.

解法1：栈Stack

一个字符串表示文件系统的目录结构，里面包含\n(回车)和\t(tab)的特殊字符，要找到最长的绝对文件路径。
可用栈Stack, Array, Hash table来保存每一层的最长路径长度。以栈为例：

1. 将字符串以'\n'进行拆分，得到目录字符串数组，设置最大长度变量 maxLen = 0，设置一个 Stack，初始设为0层为0。

2. 对这个数组迭代，对于每个目录字符串统计其前面的'\t'个数， 个数加1就是当前目录的层深度，比如： 0个'\t'，层数就是1，1个'\t'，层数是2。

3. 比较当前目录层数和栈顶层数：

    1）如果目录层数大于栈顶层数：判断是否含'.' ，

       a. 不含'.'，则是目录，计算当前目录长度（字符长度+1因为要多1个"/"）+ 上层目录最大长度，长度入栈。

       b. 含'.' , 则为文件，计算当前文件长度（字符长度），此长度加上栈顶元素长度（上层目录最大长度）和maxLen比较取最大。

    2）如果目录层数小于栈顶层数：

       则弹出栈顶元素，用while循环，一直弹到栈顶的层小于当前层数。然后重复上面的a, b步骤

最后，返回maxLen.

解法2：哈希表HashMap

Java: Stack, Time: O(n), Space: O(n)

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public class Solution {     public int lengthLongestPath(String input) {         String[] lines = input.split("\n");         Stack<Integer> path = new Stack<>();         path.push(0);         int max = 0;                   for (String line : lines) {             int tabIndex = line.lastIndexOf("\t") + 1;             while (path.size() - 1 > tabIndex) {                 path.pop();             }             int length = path.peek() + 1 + line.length() - tabIndex;             path.push(length);             if (line.indexOf(".") != -1)                 max = Math.max(max, length - 1);         }                   return max;     } }

Java: DP Array, Time: O(n), Space: O(n)

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public class Solution {     public int lengthLongestPath(String input) {         String[] lines = input.split("\n");         int[] length = new int[lines.length + 1];         int max = 0;                   for (String line : lines) {             int depth = line.lastIndexOf("\t") + 1;             length[depth + 1] = length[depth] + 1 + line.length() - depth;             if (line.indexOf(".") != -1) {                 max = Math.max(max, length[depth + 1] - 1);             }         }         return max;     } }

Java: Stack

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class Solution {     public int lengthLongestPath(String input) {         Deque<Integer> stack = new ArrayDeque<>();         stack.push(0); // "dummy" length         int maxLen = 0;         for(String s:input.split("\n")){             int lev = s.lastIndexOf("\t") + 1; // number of "\t"             while(lev + 1 < stack.size()) stack.pop(); // find parent             int len = stack.peek() + s.length() - lev + 1; // remove "/t", add"/"             stack.push(len);             // check if it is file             if(s.contains(".")) maxLen = Math.max(maxLen, len - 1);         }         return maxLen;     } }

Java: Array instead of Stack

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class Solution {     public int lengthLongestPath(String input) {         String[] paths = input.split("\n");         int[] stack = new int[paths.length+1];         int maxLen = 0;         for(String s:paths){             int lev = s.lastIndexOf("\t")+1, curLen = stack[lev+1] = stack[lev] + s.length() - lev + 1;             if(s.contains(".")) maxLen = Math.max(maxLen, curLen - 1);         }         return maxLen;     }　　 }

Python: HashMap

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class Solution(object):     def lengthLongestPath(self, input):         """         :type input: str         :rtype: int         """         maxlen = 0         pathlen = {0: 0}         for line in input.splitlines():             name = line.lstrip('\t')             depth = len(line) - len(name)             if '.' in name:                 maxlen = max(maxlen, pathlen[depth] + len(name))             else:                 pathlen[depth + 1] = pathlen[depth] + len(name) + 1         return maxlen

Python: HashMap

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class Solution(object):     def lengthLongestPath(self, input):         """         :type input: str         :rtype: int         """         def split_iter(s, tok):             start = 0             for i in xrange(len(s)):                 if s[i] == tok:                     yield s[start:i]                     start = i + 1             yield s[start:]             max_len = 0         path_len = {0: 0}         for line in split_iter(input, '\n'):             name = line.lstrip('\t')             depth = len(line) - len(name)             if '.' in name:                 max_len = max(max_len, path_len[depth] + len(name))             else:                 path_len[depth + 1] = path_len[depth] + len(name) + 1         return max_len

C++: Stack

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class Solution {  public:      int lengthLongestPath(string input) {          input+= '\n';          int len = input.size(), ans = 0, spaceCnt = 0, curPathLen =0, dotFlag = 0;          stack<pair<string, int> > st;          string curDir, extension;          for(int i = 0; i < len; i++)          {              if(input[i]!='\n')              {                  if(input[i]!='\t') curDir += input[i];                  if(dotFlag) extension += input[i];                  if(input[i]=='\t') spaceCnt++;                  if(input[i]=='.') dotFlag = 1;                  continue;              }                while(!st.empty() && spaceCnt <= st.top().second)              {                  curPathLen -= st.top().first.size();                  st.pop();              }              if(dotFlag) ans = max(curPathLen + (int)curDir.size() + 1, ans);              else              {                  st.push(make_pair("/"+ curDir, spaceCnt));                  curPathLen += curDir.size() + 1;              }              extension = "", dotFlag = 0, spaceCnt = 0, curDir = "";          }          return ans==0?0:ans-1;      }  };

　　

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