1233. Remove Sub-Folders from the Filesystem

问题：

给出一组全路径文件数组，删除子目录。

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.

　　

解法：

先对数组排序，得到的顺序则为：

父目录，在子目录前，

从头往后遍历，先加入结果的目录，一定为后加入的父目录，或者两者没有父子关系。

那么只有当前最后加入结果的目录，有可能为下一个遍历目录的父目录，

因此只用判断res.back()是否为当前遍历值的父目录即可。

这里的判断方法为：

当前遍历值.substr()==res.back()+"/"  ？？

若是，则不需要加当前遍历值入res，否则，加入。

代码参考：

class Solution {
public:
    vector<string> removeSubfolders(vector<string>& folder) {
        vector<string> res;
        sort(folder.begin(), folder.end());
        for(string cur:folder){
            if(!res.empty()){
                string tmp=res.back()+"/";
                int len=tmp.size();
                if(cur.substr(0,len)==tmp) continue;  
            }
            res.push_back(cur);
        }
        return res;
    }
};