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，首先應該想到使用Trie這種資料結構的可能性，這道題涉及到字尾問題，所以我們需要設計一個函式rinsert來從單詞最後一個字元開始建立Trie（insert函式和rinsert函式的差別僅僅在於遍歷字串的順序相反而已）。因此tree是字首樹，rtree就成了字尾樹。在TrieNode中，我還維護了一個last陣列用於記錄走過這個節點的索引值。我們通過訪問一個TrieNode可以得到所有以該字首在words中的索引號，即startWith的函式功能。所以，我們只需要所有出現過prefix的陣列pre和所有出現過suffix的陣列suf中找到同時出現且最大的索引號即可。

Code:

 1 struct TrieNode{
 2     TrieNode *next[26];
 3     bool isEnd;
 4     vector<int> last;
 5     TrieNode():isEnd(false),last(0){
 6         for(int i = 0;i != 26;++i)
 7             next[i] = NULL;
 8     }
 9 };
10 class TrieTree{
11 public:
12     TrieTree(){
13         root = new
 
 TrieNode();
14     }
15     void insert(string word,int index) {
16         TrieNode *current = root;
17         current->last.push_back(index);
18         for(int i = 0;i != word.size();++i){
19             if(current->next[word[i]-'a'] == NULL)
20                 current->next[word[i]-'a'] = new TrieNode();
21             current = current->next[word[i]-'a'];
22             current->last.push_back(index);
23         }
24         current->isEnd = true;
25     }
26     void rinsert(string word,int index) {
27         TrieNode *current = root;
28         current->last.push_back(index);
29         for(int i = word.size()-1;i >= 0;--i){
30             if(current->next[word[i]-'a'] == NULL)
31                 current->next[word[i]-'a'] = new TrieNode();
32             current = current->next[word[i]-'a'];
33             current->last.push_back(index);
34         }
35         current->isEnd = true;
36     }
37     vector<int> startWith(string prefix){
38         TrieNode *current = root;
39         for(int i = 0;i != prefix.size();++i){
40             if(current->next[prefix[i]-'a'] == NULL)
41                 return {};
42             current = current->next[prefix[i]-'a'];
43         }
44         return current->last;
45     }
46     struct TrieNode *getRoot(){
47         return root;
48     }
49 private:
50     TrieNode *root;
51 };
52 class WordFilter {
53 public:
54     WordFilter(vector<string> words) {
55         for(int i = 0;i != words.size();++i){
56             tree.insert(words[i],i);
57             rtree.rinsert(words[i],i);
58         }
59         
60     }
61     
62     int f(string prefix, string suffix) {
63         vector<int> pre = tree.startWith(prefix);
64         reverse(suffix.begin(),suffix.end());
65         vector<int> suf = rtree.startWith(suffix);
66         for(int i = pre.size()-1;i >= 0;--i){
67             for(int j = suf.size()-1;j >= 0;--j){
68                 if(pre[i] == suf[j]) 
69                     return pre[i];
70                 else if(pre[i] > suf[j])
71                     break;
72             }
73         }
74         return -1;
75     }
76 private:
77     TrieTree tree;
78     TrieTree rtree;
79 };
80 
81 /**
82  * Your WordFilter object will be instantiated and called as such:
83  * WordFilter obj = new WordFilter(words);
84  * int param_1 = obj.f(prefix,suffix);
85  */