1233. Remove Sub-Folders from the Filesystem**

1233. Remove Sub-Folders from the Filesystem**

https://leetcode.com/problems/remove-sub-folders-from-the-filesystem/

题目描述

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 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 * 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.

C++ 实现 1

这道题需要注意观察, 容易想到的思路是对 folder 列表按照字典序进行排序, 然后观察 folder[i] 和 folder[i + 1], ... 之间的关系, 是否 folder[i + 1], ... 之类以 folder[i] 开头, 用 Python 里面的字符串函数来说就是 if folder[i + 1].startswith(folder[i]) 成立, 如果成立, 那么 folder[i + 1] 就不用考虑了.

但需要注意的是, 观察上面第 3 个例子, 对于 "/a/b/ca" 和 "/a/b/c", 就会发现上面的思路有缺陷. 为了解决这个问题, 可以对这些路径末尾都加上一个 /, 即变成了 "/a/b/ca/" 和 "/a/b/c/" 的对比. 然而这要求原路径本身末尾不包含 /. 根据题目描述:

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.

路径以 / 开头, 以 lowercase English letters 结尾, 那么末尾应该不包含 /.

我的解法, 有点繁琐. 可以看 C++ 实现 2, 代码更简洁.

class Solution {
private:
    bool startswith(const string &a, const string &b) {
        int k = 0;
        while (k < a.size() && k < b.size()) {
            if (a[k] != b[k]) return false;
            ++ k;
        }
        return true;
    }
public:
    vector<string> removeSubfolders(vector<string>& folder) {
        for (auto &s : folder) {
            if (s.back() != '/')
                s += '/';
        }
        std::sort(folder.begin(), folder.end());
        vector<string> res;
        for (int i = 0; i < folder.size(); ++ i) {
            auto parent_folder = folder[i];
            res.push_back(folder[i].substr(0, folder[i].size() - 1));
            while (i + 1 < folder.size() && startswith(folder[i + 1], parent_folder))
                ++ i;
        }
        return res;
    }
};

C++ 实现 2

class Solution {
public:
    vector<string> removeSubfolders(vector<string>& folder) {
        std::sort(folder.begin(), folder.end());
        vector<string> res;
        for (int i = 0; i < folder.size(); ++ i) {
            auto prefix = folder[i] + '/';
            res.push_back(folder[i]);
			// 利用 folder[i + 1].substr(0, prefix.size()) == prefix
			// 实现类似 folder[i + 1].startswith(prefix) 的效果
            while (i + 1 < folder.size() && folder[i + 1].substr(0, prefix.size()) == prefix)
                ++ i;
        }
        return res;
    }
};