字典树Trie实战

问起起源于在学堂在线上的《软件工程》这门课，一般每章后面都会有一个实践作业，发现其难度可以比拟大一大二的课程大作业，早有这样的训练强度该有多好呢？呵呵O(∩_∩)O~

A. 问题描述

以上是基础的要求，其实字符串的题目一般都可以随便暴力做，所以评价一个算法的优劣，主要是看它在特定应用场景下的效率，比如：



基于上面的要求，我决定使用“字典树”Trie来实现基于字符串的快速查找，字典树的描述请自行搜索，这里暂时只分享我自己的python 3实现。

B. 实现

初次实现，有点乱，先mark一下，过些日子回来完善【未完待续】

# !/usr/bin/env python3
# -*- coding: utf-8 -*-
__author__ = 'jacket'

import sys
import re

SIZE = 27

def char2int(ch):
# return ord(ch) - ord('a') + 1
# for quick calculation, ord('a') = 97, so the above equal:
return ord(ch) - 96

class TrieNode:
def __init__(self, data=None):
self.data = data
self.cnt = 0
self.leafNode = None
self.children = []

def haveChildren(self):
return len(self.children) == SIZE

def extend(self):
self.children = [TrieNode() for _ in range(SIZE)]

def addToLeaf(self, word):
if not self.leafNode:
self.leafNode = TrieNode(word)
self.leafNode.cnt += 1
return self.leafNode.cnt - 1

def addToChild(self, ch, index):
if not self.children[index].data:
self.children[index].data = ch
self.children[index].cnt += 1

if not self.leafNode:
self.leafNode = TrieNode()
self.leafNode.cnt += 1

class Trie:
def __init__(self):
self.root = TrieNode()

def insert(self, word):
now = self.root

for ch in word:
index = char2int(ch)
if not now.haveChildren():
now.extend()
now.addToChild(ch, index)
now = now.children[index]

return now.addToLeaf(word)

def main(args):
t = Trie()
with open(args[0], 'r') as f:
for line in f:
if line[-1] == '\n':
line = line[0:-1]
if len(line) > 0
t.insert(line.lower())

print('Trie build')
while True:
q = input()
if q == '0':
break
print(t.insert(q.lower()))

return 0

if __name__ == '__main__':
exit(main(sys.argv[1:]))

C. More

图片来自wiki百科