Leetcode 745. Prefix and Suffix Search

Problem:

Given many words, words[i] has weight i.

Design a class WordFilter that supports one function, WordFilter.f(String prefix, String suffix). It will return the word with given prefix and suffix with maximum weight. If no word exists, return -1.

Examples:

Input:
WordFilter(["apple"])
WordFilter.f("a", "e") // returns 0
WordFilter.f("b", "") // returns -1

Note:

1. words has length in range [1, 15000].
2. For each test case, up to words.length queries WordFilter.f may be made.
3. words[i] has length in range [1, 10].
4. prefix, suffix have lengths in range [0, 10].
5. words[i] and prefix, suffix queries consist of lowercase letters only.

Solution:

　　看到prefix和suffix，首先应该想到使用Trie这种数据结构的可能性，这道题涉及到后缀问题，所以我们需要设计一个函数rinsert来从单词最后一个字符开始创建Trie（insert函数和rinsert函数的差别仅仅在于遍历字符串的顺序相反而已）。因此tree是前缀树，rtree就成了后缀树。在TrieNode中，我还维护了一个last数组用于记录走过这个节点的索引值。我们通过访问一个TrieNode可以得到所有以该前缀在words中的索引号，即startWith的函数功能。所以，我们只需要所有出现过prefix的数组pre和所有出现过suffix的数组suf中找到同时出现且最大的索引号即可。

Code:

 

struct TrieNode{
    TrieNode *next[26];
    bool isEnd;
    vector<int> last;
    TrieNode():isEnd(false),last(0){
        for(int i = 0;i != 26;++i)
            next[i] = NULL;
    }
};
class TrieTree{
public:
    TrieTree(){
        root = new TrieNode();
    }
    void insert(string word,int index) {
        TrieNode *current = root;
        current->last.push_back(index);
        for(int i = 0;i != word.size();++i){
            if(current->next[word[i]-'a'] == NULL)
                current->next[word[i]-'a'] = new TrieNode();
            current = current->next[word[i]-'a'];
            current->last.push_back(index);
        }
        current->isEnd = true;
    }
    void rinsert(string word,int index) {
        TrieNode *current = root;
        current->last.push_back(index);
        for(int i = word.size()-1;i >= 0;--i){
            if(current->next[word[i]-'a'] == NULL)
                current->next[word[i]-'a'] = new TrieNode();
            current = current->next[word[i]-'a'];
            current->last.push_back(index);
        }
        current->isEnd = true;
    }
    vector<int> startWith(string prefix){
        TrieNode *current = root;
        for(int i = 0;i != prefix.size();++i){
            if(current->next[prefix[i]-'a'] == NULL)
                return {};
            current = current->next[prefix[i]-'a'];
        }
        return current->last;
    }
    struct TrieNode *getRoot(){
        return root;
    }
private:
    TrieNode *root;
};
class WordFilter {
public:
    WordFilter(vector<string> words) {
        for(int i = 0;i != words.size();++i){
            tree.insert(words[i],i);
            rtree.rinsert(words[i],i);
        }
        
    }
    
    int f(string prefix, string suffix) {
        vector<int> pre = tree.startWith(prefix);
        reverse(suffix.begin(),suffix.end());
        vector<int> suf = rtree.startWith(suffix);
        for(int i = pre.size()-1;i >= 0;--i){
            for(int j = suf.size()-1;j >= 0;--j){
                if(pre[i] == suf[j]) 
                    return pre[i];
                else if(pre[i] > suf[j])
                    break;
            }
        }
        return -1;
    }
private:
    TrieTree tree;
    TrieTree rtree;
};

/**
 * Your WordFilter object will be instantiated and called as such:
 * WordFilter obj = new WordFilter(words);
 * int param_1 = obj.f(prefix,suffix);
 */