删除子文件夹

1. 问题描述
题目链接：删除子文件夹

2. 思路
看到公共前缀，考虑用字典树，对于该题，需要对每个节点额外开辟一个标志位end，以标记是否为完整路径结束。
字典树结构如下

点击查看代码

struct Node {
    int end;
    unordered_map<string, Node*> next;
    Node() {
        end = 0;
    };
};

class Trie {
public:
    Trie() {
        root = new Node();
    };
    void insert(vector<string>& words) {};
    bool search(vector<string>& words) {};
private:
    Node* root;
};

3. 完整代码

点击查看代码

struct Node {
    int end;
    unordered_map<string, Node*> next;
    Node() {
        end = 0;
    };
};

class Trie {
public:
    Trie() {
        root = new Node();
    };
    void insert(vector<string>& words) {
        Node* cur = root;
        for (auto word : words) {
            if (cur->next.find(word) == cur->next.end()) {
                cur->next[word] = new Node();
            }
            cur = cur->next[word];
        }
        cur->end = 1;
    }
    bool search(vector<string>& words) {
        Node* cur = root;
        for(int i = 0; i < words.size(); i++) {
            cur = cur->next[words[i]];
            if(i < words.size() - 1 && cur->end) return true;        // 前缀不能是word数组本身
        }
        return false;
    }
private:
    Node* root;
};

class Solution {
public:
    vector<string> split(string& s, char delim) {
        stringstream ss(s);
        string item;
        vector<string> res;
        while (getline(ss, item, delim)) {
            res.emplace_back(item);
        }
        return res;
    }
    vector<string> removeSubfolders(vector<string>& folder) {
        int n = folder.size();
        vector<int> flag(n);
        Trie* trie = new Trie();
        vector<vector<string>> paths;
        for (int i = 0; i < n; ++i) {
            vector<string> path = split(folder[i], '/');
            paths.push_back(path);
            trie->insert(path);
        }
        for (int i = 0; i < n; ++i) {
            if (trie->search(paths[i])) {
                flag[i] = 1;
            }
        }
        vector<string> ans;
        for(int i = 0; i < n; i++) {
            if(!flag[i]) ans.push_back(folder[i]);
        }
        return ans;
    }
};

4. 执行结果