题目如下:
Suppose we abstract our file system by a string in the following manner:
The string
"dir subdir1 subdir2 file.ext"
represents:dir subdir1 subdir2 file.extThe directory
dir
contains an empty sub-directorysubdir1
and a sub-directorysubdir2
containing a filefile.ext
.The string
"dir subdir1 file1.ext subsubdir1 subdir2 subsubdir2 file2.ext"
represents:dir subdir1 file1.ext subsubdir1 subdir2 subsubdir2 file2.extThe directory
dir
contains two sub-directoriessubdir1
andsubdir2
.subdir1
contains a filefile1.ext
and an empty second-level sub-directorysubsubdir1
.subdir2
contains a second-level sub-directorysubsubdir2
containing a filefile2.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 is32
(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)
wheren
is the size of the input string.Notice that
a/aa/aaa/file1.txt
is not the longest file path, if there is another pathaaaaaaaaaaaaaaaaaaaaa/sth.png
.
解题思路:我的方法很简单,首先把input按' '分割,接下来判断分割后的每一个子串前缀有几个' ',并以' '的个数作为key值将子串存入字典中,而这个子串的上一级目录是(key-1)在字典中的最后一个元素。遍历所有的子串后即可求得最大长度。
代码如下:
class Solution(object): def lengthLongestPath(self, input): """ :type input: str :rtype: int """ #print input dic = {} dic[0] = [''] res = 0 if input.count('.') < 1: return 0 itemlist = input.split(' ') for i in itemlist: count = i.count(' ') + 1 if count not in dic: dic[count] = [dic[count-1][-1] + '/' + i.replace(' ','')] else: dic[count].append(dic[count-1][-1]+ '/'+i.replace(' ','')) #print len(dic[count][-1])-1,dic[count][-1] if dic[count][-1].count('.') >= 1: res = max(res,len(dic[count][-1])-1) #print dic return res