单词替换

题目表述

在英语中，我们有一个叫做 词根(root) 的概念，可以词根后面添加其他一些词组成另一个较长的单词——我们称这个词为 继承词(successor)。例如，词根an，跟随着单词 other(其他)，可以形成新的单词 another(另一个)。

现在，给定一个由许多词根组成的词典 dictionary 和一个用空格分隔单词形成的句子 sentence。你需要将句子中的所有继承词用词根替换掉。如果继承词有许多可以形成它的词根，则用最短的词根替换它。

你需要输出替换之后的句子。

示例 1

输入：dictionary = ["cat","bat","rat"], sentence = "the cattle was rattled by the battery"

输出："the cat was rat by the bat"

示例 2

输入：dictionary = ["a","b","c"], sentence = "aadsfasf absbs bbab cadsfafs"

输出："a a b c"

提示

1 <= dictionary.length <= 1000

1 <= dictionary[i].length <= 100

dictionary[i] 仅由小写字母组成。

1 <= sentence.length <= 10^6

sentence 仅由小写字母和空格组成。

sentence 中单词的总量在范围 [1, 1000] 内。

sentence 中每个单词的长度在范围 [1, 1000] 内。

sentence 中单词之间由一个空格隔开。

sentence 没有前导或尾随空格。

解法一：使用自带的startswith函数判断词根

遍历句子中的每个单词，对于每个词根，使用自带的startswith函数判断词根是不是存在与单词中。

时间复杂度为 O(mn)，空间复杂度为 O(n)

代码实现

from typing import List


class Solution:
    """方案一：使用自带的startswith函数判断词根"""
    def replaceWords(self, dictionary: List[str], sentence: str) -> str:
        sentence_list = sentence.split(" ")
        dictionary_sorted = sorted(dictionary, key=lambda word:len(word))
        for i, word in enumerate(sentence_list):
            for root in dictionary_sorted:
                # 找到词根，替换
                if word.startswith(root):
                    sentence_list[i] = root
                    break
        return " ".join(sentence_list)

package leetcode

import (
	"sort"
	"strings"
)

// 方案一：使用自带的startswith函数判断词根
func replaceWords(dictionary []string, sentence string) string {
	sentenceWords := strings.Split(sentence, " ")
	sort.Slice(dictionary, func(i, j int) bool {
		return len(dictionary[i]) < len(dictionary[j])
	})
	for i, word := range sentenceWords {
		for _, root := range dictionary {
			if strings.HasPrefix(word, root) {
				sentenceWords[i] = root
				break
			}
		}
	}
	return strings.Join(sentenceWords, " ")
}

解法二：hash表

遍历句子中的每个单词，遍历每个单词的字母序，使用hash表记录词根是否存在。

时间复杂度：O(mn)，空间复杂度：O(n)

代码实现

from typing import List

class Solution2:
    """方案二：使用 hash 表判断"""
    def replaceWords(self, dictionary: List[str], sentence: str) -> str:
        words = sentence.split(" ")
        dictionary_set = set(dictionary)
        for i, word in enumerate(words):
            for j in range(1, len(words)+1):
                if word[:j] in dictionary_set:
                    # 找到词根，替换
                    words[i] = word[:j]
                    break
        return " ".join(words)

package leetcode

import "strings"

// 方案二：使用 hash 表判断
func replaceWords2(dictionary []string, sentence string) string {
	dictionarySet := map[string]bool{}
	for _, root := range dictionary {
		dictionarySet[root] = true
	}
	words := strings.Split(sentence, " ")
	for i, word := range words {
		for j := 1; j <= len(word); j++ {
			if dictionarySet[word[:j]] {
				words[i] = word[:j]
				break
			}
		}
	}
	return strings.Join(words, " ")
}

解法二：hash表

遍历句子中的每个单词，遍历每个单词的字母序，使用hash表记录词根是否存在。

时间复杂度：O(max(m,n))，空间复杂度：O(m)

代码实现

from typing import List

class Solution2:
    """方案二：使用 hash 表判断"""
    def replaceWords(self, dictionary: List[str], sentence: str) -> str:
        words = sentence.split(" ")
        dictionary_set = set(dictionary)
        for i, word in enumerate(words):
            for j in range(1, len(words)+1):
                if word[:j] in dictionary_set:
                    # 找到词根，替换
                    words[i] = word[:j]
                    break
        return " ".join(words)

package leetcode

import "strings"

// 方案二：使用 hash 表判断
func replaceWords2(dictionary []string, sentence string) string {
	dictionarySet := map[string]bool{}
	for _, root := range dictionary {
		dictionarySet[root] = true
	}
	words := strings.Split(sentence, " ")
	for i, word := range words {
		for j := 1; j <= len(word); j++ {
			if dictionarySet[word[:j]] {
				words[i] = word[:j]
				break
			}
		}
	}
	return strings.Join(words, " ")
}

解法三：字典树

构建一颗字典树，字典树的内容是词根列表。

遍历句子，对于每个单词打断是否存在于字典树中

时间复杂度：O(max(n,m))，空间复杂度：O(m)

代码实现

from typing import List

class Solution:
    """方案三：字典树"""
    def replaceWords(self, dictionary: List[str], sentence: str) -> str:
        # 使用词根构建字典树        
        trie = {}
        for word in dictionary:
            cur = trie
            for c in word:
                if c not in cur:
                    cur[c] = {}
                cur = cur[c]
            cur["#"] = {}
        words = sentence.split(' ')
        for i, word in enumerate(words):
            cur = trie
            for j, c in enumerate(word):
                if '#' in cur:
                    words[i] = word[:j]
                    break
                if c not in cur:
                    break
                cur = cur[c]
        return " ".join(words)

package leetcode

import "strings"

// 方案三：字典树
func replaceWords3(dictionary []string, sentence string) string {
	type trie map[rune]trie

	root := trie{}
	for _, s := range dictionary {
		cur := root
		for _, c := range s {
			if cur[c] == nil {
				cur[c] = trie{}
			}
			cur = cur[c]
		}
		cur['#'] = trie{}
	}

	words := strings.Split(sentence, " ")
	for i, word := range words {
		cur := root
		for j, c := range word {
			if cur['#'] != nil {
				words[i] = word[:j]
				break
			}
			if cur[c] == nil {
				break
			}
			cur = cur[c]
		}
	}
	return strings.Join(words, " ")
}