745. Prefix and Suffix Search

问题描述：

 

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.

解题思路：

看到根据前缀找相应的字符串，首先想到了trie树。

由于要返回最大的权重的字符串，我们可以用一个map来存储字符串和相应的权重。

这里要实现的方法是：根据前缀返回包含此前缀的所有的字符串， 可以用DFS来进行寻找。

对包含次前缀的字符串，我们再对其后缀进行查验，并且检查其对应权重，要返回权重最大的字符串

代码：

class TrieNode {
public:
    TrieNode* child[26];
    bool isWord;
    TrieNode(): isWord(false){
        for(auto &a : child){
            a = NULL;
        }
    }
};
class WordFilter {
public:
    WordFilter(vector<string> words) {
        root = new TrieNode();
        int i = 0;
        for(auto w : words){
            insert(w);
            weight_map[w] = i++;
        }
    }
    
    int f(string prefix, string suffix) {
        vector<string> pre = findAllPrefix(prefix);
        if(pre.empty()) return -1;
        int weight = -1;
        for(auto w : pre){
            if(isSufix(w, suffix) && weight_map[w] > weight){
                weight = weight_map[w];
            }
        }
        return weight;
    }
private:
    unordered_map<string, int> weight_map;
    TrieNode* root;
    
    void insert(string word){
        TrieNode* p = root;
        for(auto c : word){
            int i = c - 'a';
            if(!p->child[i]){
                p->child[i] = new TrieNode();
            }
            p = p->child[i];
        }
        p->isWord = true;
    }
    
    bool isSufix(string w, string suf){
        int m = w.size()-1;
        int n = suf.size()-1;
        if(m < n) return false;
        while(n > -1){
            if(suf[n--] != w[m--]) return false;
        }
        return true;
    }
    void findAllWords(vector<string> &allW, TrieNode* p, string cur){
        if(p->isWord) allW.push_back(cur);
        for(int i = 0; i < 26; i++){
            if(p->child[i]){
                cur.push_back(char('a'+i));
                findAllWords(allW, p->child[i], cur);
                cur.pop_back();
            }
        }
    }
    
    vector<string> findAllPrefix(string pre){
        TrieNode* p = root;
        vector<string> ret;
        for(auto c : pre){
            int i = c - 'a';
            if(!p->child[i]) return ret;
            p = p->child[i];
        }
        findAllWords(ret, p, pre);
        return ret;
    }
    
};

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