【手把手带你刷Leetcode力扣】11.数据结构 - 字典树/前缀树Trie

本文链接：https://blog.youkuaiyun.com/ZZBOOM_/article/details/120839868

本文探讨了前缀树（Trie）在字符串匹配问题中的高效解决方案，对比了传统数组方法的效率。通过示例展示了如何使用前缀树进行插入、搜索和前缀检查操作，以及在寻找词典中最长单词的问题上应用前缀树的方法。代码实现包括暴力法和前缀树法，两种方法的时间复杂度分析。视频教程来源于B站—爱学习的饲养员—手把手教你刷Leetcode力扣。

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字典树/前缀树Trie

自动匹配的程序
数组：[goog, googl, google, bai, baidu, gi]

goog
googl
google
gi

goog
googl
google

插入：O（N）
搜索：O（N）
前缀存在：O（NM）(元素个数匹配的时间复杂度)

Trie
root节点 root = Trie()
孩子节点 Hash
结束Flag
对应的值

插入：O（K）
搜索：O（K）
前缀存在：O（K）

词频统计

词典中最长的单词
暴力法：

class Solution:
	# N is the size of words, M is the size of each word in words
	# Time Complexity: O(sigma(M*N))
	# Space Complexity: O(sigma(M*N))
	def longestWord(self, word: List[str]) -> str:
		if words is None or len(words) == 0:
			return ""
		result = ""
		hashset = set(words)
		for word in words:
			if len(word) > len(result) or (len(word) == len(result) and word < result):
				isWord = True
				l = len(word)
				for i in range(0, l):
					if word[0:i+1] not in hashset:
						isWord = False
						break
				result = word if isWord else result
		return result

前缀树：

class Solution:
	# N is the size of words, M is the size of each word in words
	# Time Complexity: O(sigma(M*N))
	# Space Complexity: O(sigma(M*N))
	def longestWord(self, words: List[str]) -> str:
		if words is None or len(words) == 0:
			return ""
		root = Trie()
		# Insert each word into trie
		for word in words:
			cur = root
			for c in word:
				if c in cur.children:
					cur = cur.children[c]
				else:
					newNode = Trie()
					cur.children[c] = newNode
					cur = newNode
			cul.val = word
			cur.isEnd = True
		# Looking for the eligible word
		result = ""
		for word in words:
			cur = root
			if len(word) > len(result) or (len(word) == len(result) and word < reult):
				isWord = True
				for c in word:
					cur = cur.children[c]
					if not cur.isEnd:
						isWord = False
						break
				result = word if isWord else result
		return result

class Trie:
	def __init__(self):
		self.children = {}
		self.isEnd = False
		self.val = ""