【leetcode】388. Longest Absolute File Path

题目如下：

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.

解题思路：我的方法很简单，首先把input按'\n'分割，接下来判断分割后的每一个子串前缀有几个'\t'，并以'\t'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('\n')
        for i in itemlist:
            count = i.count('\t') + 1
            if count not in dic:
                dic[count] = [dic[count-1][-1] + '/' + i.replace('\t','')]
            else:
                dic[count].append(dic[count-1][-1]+ '/'+i.replace('\t',''))
            #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