Who's Studio.

hihoCoder#1014 - Trie树

Word count: 479Reading time: 2 min
2016/05/11

题目

Trie树就是字典树。对于一个给定的词典,任给一个字符串,输出词典中以该字符串为前缀的单词数目。
具体描述请见hihoCoder

解题思路

在建立Trie树的同时我们就可以着手统计以某个字符串为前缀的节点个数,即每次添加一个新单词的时候,沿途更新所有经过的节点。

时间复杂度

单词查询用时26M(M:查询字符串的长度),单次插入用时26M。

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
#include <iostream>
#include <string>
#include <vector>
using namespace std;

struct Node {
int L = 1;
char ch = ' ';
vector<Node*> successors;

Node(char c) {
ch = c;
}
};

class Trie {
private:
Node* root;

void updatePrefix(Node* node, string word, int idx) {
if (idx == word.size())
return;

bool find = false;
for (int i = 0; i < node->successors.size(); i++) {
if (node->successors[i]->ch == word[idx]) {
find = true;
node->successors[i]->L++;
updatePrefix(node->successors[i], word, idx + 1);
break;
}
}

if (!find) {
for (int i = idx; i < word.size(); i++) {
node->successors.push_back(new Node(word[i]));
node = node->successors.back();
}
}
}

int findPrefix(Node* node, string query, int idx) {
for (int i = 0; i < node->successors.size(); i++) {
if (node->successors[i]->ch == query[idx]) {
if (idx == query.size() - 1)
return node->successors[i]->L;

return findPrefix(node->successors[i], query, idx + 1);
}
}

return 0;
}

void deleteTrie(Node* node) {
for (int i = 0; i < node->successors.size(); i++)
deleteTrie(node->successors[i]);
delete node;
}

public:
Trie() {
root = new Node('$');
}

~Trie() {
deleteTrie(root);
}

void insert(string word) {
updatePrefix(root, word, 0);
}

int getStat(string query) {
return findPrefix(root, query, 0);
}
};

int main() {
int n = 0;
cin >> n;
string* words = new string[n];
for (int i = 0; i < n; i++)
cin >> words[i];
int m = 0;
cin >> m;
string* queries = new string[m];
for (int i = 0; i < m; i++)
cin >> queries[i];

Trie trie;
for (int i = 0; i < n; i++)
trie.insert(words[i]);

for (int i = 0; i < m; i++)
cout << trie.getStat(queries[i]) << endl;

delete[] words, queries;

return 0;
}
CATALOG
  1. 1. 题目
  2. 2. 解题思路
  3. 3. 时间复杂度
  4. 4. 代码