牛客——接头密匙

该题是一个很显然的前缀树问题，只需要构建a中所有数组对应的前缀树，之后求b所处前缀个数即可。关于前缀树的构建，可以观看左老师算法讲解045的视频，简单来讲就是用特殊字符将实际数据隔开，同时将实际数据离散化。示例题解如下：

cpp 复制代码

class Solution {
  private:
    static constexpr int N = 1e6 + 1;
    vector<vector<int>> trie = vector<vector<int>>(N,
                               vector<int>(12));   // 0 - 9 & '-' & '#'
    vector<int> passArr = vector<int>(N);
    vector<int> endArr = vector<int>(N);
    int cnt;

    int getPath(char c) {
        if (c == '-') {
            return 10;
        }
        if (c == '#') {
            return 11;
        }
        return c - '0';
    }

    void buildTrie() {
        cnt = 1;
    }

    void reBuildTrie() {
        for (int i = 0; i < cnt; ++i) {
            fill(&trie[i][0], &trie[i][0] + 12, 0);
            passArr[i] = 0;
            endArr[i] = 0;
        }
    }

    void insert(const string& word) {
        int cur = 1, path;
        passArr[cur]++;
        for (char c : word) {
            path = getPath(c);
            if (trie[cur][path] == 0) {
                trie[cur][path] = ++cnt;
            }
            cur = trie[cur][path];
            passArr[cur]++;
        }
        endArr[cur]++;
    }

    int prefixCount(const string& word) {
        int cur = 1, path;
        for (char c : word) {
            path = getPath(c);
            if (trie[cur][path] == 0) {
                return 0;
            }
            cur = trie[cur][path];
        }
        return passArr[cur];
    }

  public:
    vector<int> countConsistentKeys(vector<vector<int> >& b,
                                    vector<vector<int> >& a) {
        // 建立前缀树
        buildTrie();
        for (int i = 0; i < a.size(); ++i) {
            // a[i] = vector<int>
            string s = "";
            for (int j = 1; j < a[i].size(); j++) {
                s = s + std::to_string(a[i][j] - a[i][j - 1]) + "#";
            }
            insert(s);
        }

        vector<int> ans;
        // 搜索前缀树
        for (int i = 0; i < b.size(); ++i) {
            // b[i] = vector<int>
            string pre = "";
            for (int j = 1; j < b[i].size(); ++j) {
                pre = pre + std::to_string(b[i][j] - b[i][j - 1]) + "#";
            }
            ans.push_back(prefixCount(pre));
        }

        reBuildTrie();

        return ans;
    }
};