【leetcode】1233. Remove Sub-Folders from the Filesystem
题目如下:
Given a list of folders, remove all sub-folders in those folders and return in any order the folders after removing.
If a
folder[i]
is located within anotherfolder[j]
, it is called a sub-folder of it.The format of a path is one or more concatenated strings of the form:
/
followed by one or more lowercase English letters. For example,/leetcode
and/leetcode/problems
are valid paths while an empty string and/
are not.Example 1:
Input: folder = ["/a","/a/b","/c/d","/c/d/e","/c/f"] Output: ["/a","/c/d","/c/f"] Explanation: Folders "/a/b/" is a subfolder of "/a" and "/c/d/e" is inside of folder "/c/d" in our filesystem.Example 2:
Input: folder = ["/a","/a/b/c","/a/b/d"] Output: ["/a"] Explanation: Folders "/a/b/c" and "/a/b/d/" will be removed because they are subfolders of "/a".Example 3:
Input: folder = ["/a/b/c","/a/b/ca","/a/b/d"] Output: ["/a/b/c","/a/b/ca","/a/b/d"]Constraints:
1 <= folder.length <= 4 * 10^4
2 <= folder[i].length <= 100
folder[i]
contains only lowercase letters and '/'folder[i]
always starts with character '/'- Each folder name is unique.
解题思路:首先建立字典树,并把folder按元素从短到长排好序,然后遍历folder,并与字典树中已有元素做前缀匹配。如果folder[i]存在于字典树中,表示是其他目录的子目录,删除;如果不存在,则把该目录插入到字典树中。
代码如下:
class TreeNode(object): def __init__(self, x): self.val = x self.childDir = {} self.isDir = False class Trie(object): dic = {} def __init__(self): """ Initialize your data structure here. """ self.root = TreeNode(None) self.dic = {} def insert(self,word): node = self.root for i in word: if i not in node.childDir: node.childDir[i] = TreeNode(i) node = node.childDir[i] node.isDir = True def isDelete(self,dir): node = self.root for i in dir: if i in node.childDir: node = node.childDir[i] if node.isDir == True: return True else: return False return False class Solution(object): def removeSubfolders(self, folder): """ :type folder: List[str] :rtype: List[str] """ folder.sort(cmp=lambda x1,x2:len(x1) - len(x2)) trie = Trie() res = [] for f in folder: if trie.isDelete(f) == False: res.append(f) trie.insert(f + '/') return res