原文链接: https://leetcode-cn.com/problems/remove-sub-folders-from-the-filesystem

英文原文

Given a list of folders folder, return the folders after removing all sub-folders in those folders. You may return the answer in any order.

If a folder[i] is located within another folder[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 * 104
• 2 <= folder[i].length <= 100
• folder[i] contains only lowercase letters and '/'.
• folder[i] always starts with the character '/'.
• Each folder name is unique.

中文题目

你是一位系统管理员，手里有一份文件夹列表 folder，你的任务是要删除该列表中的所有 子文件夹，并以 任意顺序 返回剩下的文件夹。

我们这样定义「子文件夹」：

• 如果文件夹 folder[i] 位于另一个文件夹 folder[j] 下，那么 folder[i] 就是 folder[j] 的子文件夹。

文件夹的「路径」是由一个或多个按以下格式串联形成的字符串：

• / 后跟一个或者多个小写英文字母。

例如，/leetcode 和 /leetcode/problems 都是有效的路径，而空字符串和 / 不是。

 

示例 1：

输入：folder = ["/a","/a/b","/c/d","/c/d/e","/c/f"]
输出：["/a","/c/d","/c/f"]
解释："/a/b/" 是 "/a" 的子文件夹，而 "/c/d/e" 是 "/c/d" 的子文件夹。

示例 2：

输入：folder = ["/a","/a/b/c","/a/b/d"]
输出：["/a"]
解释：文件夹 "/a/b/c" 和 "/a/b/d/" 都会被删除，因为它们都是 "/a" 的子文件夹。

示例 3：

输入：folder = ["/a/b/c","/a/b/d","/a/b/ca"]
输出：["/a/b/c","/a/b/ca","/a/b/d"]

 

提示：

• 1 <= folder.length <= 4 * 10^4
• 2 <= folder[i].length <= 100
• folder[i] 只包含小写字母和 /
• folder[i] 总是以字符 / 起始
• 每个文件夹名都是唯一的

通过代码

高赞题解

思路：父文件夹是子文件夹的前缀，所以排序后父文件夹一定在子文件前。
因此用一个set记录父文件夹，再比较其后的文件夹的前缀是否有它即可。

class Solution {
public:
    vector<string> removeSubfolders(vector<string>& folder) {
        sort(folder.begin(),folder.end());
        unordered_set<string> father;
        vector<string> ans;
        for(string s:folder){
            string f="/";
            bool is=false;      //is表示是否为子文件夹
            for(int i=1;i<s.size();++i){
                if(s[i]=='/'){
                    if(father.count(f)){ //若前缀在p中则说明z是子文件夹
                        is=true;
                        break;
                    }
                    f+=s[i];
                }
                else{
                    f+=s[i];
                }
            }
            if(is==false){
                father.insert(s);
                ans.push_back(s);
            }
        }
        return ans;
    }
};

改进：
又因为排序后，父文件夹和其子文件夹必定是连续排列的，所以若一个文件夹不是上一个父文件夹的子文件夹，那它自己就是一个父文件夹

class Solution {
public:
    vector<string> removeSubfolders(vector<string>& folder) {
        sort(folder.begin(),folder.end());
        vector<string> ans;
        ans.push_back(folder[0]);
        for(int i=1;i<folder.size();++i){
            string father=ans.back()+"/",cur=folder[i]; //这里father+"/"是为了避免 将 /a/bc 视为 /a/b的子文件夹
            if(cur.find(father)==cur.npos){
                ans.push_back(cur);
            }
        }
        return ans;
    }
};

统计信息

通过次数	提交次数	AC比率
8469	17429	48.6%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言