472. Concatenated Words
選択した UI 言語に合わせて問題文をロシア語から翻訳します。コードは変更しません。
Дан 配列 строк words (без дубликатов). return все составные слова из данного списка слов.
Составное слово определяется как 文字列, которая полностью состоит как минимум из двух более коротких слов (не обязательно различных) из данного 配列а.
例:
Input: words = ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]
Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]
Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats";
"dogcatsdog" can be concatenated by "dog", "cats" and "dog";
"ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat".
C# 解法
照合済み/オリジナルusing System;
using System.Collections.Generic;
public class Solution {
private bool Dfs(string word, int length, bool[] visited, HashSet<string> dictionary) {
if (length == word.Length) {
return true;
}
if (visited[length]) {
return false;
}
visited[length] = true;
for (int i = word.Length - (length == 0 ? 1 : 0); i > length; --i) {
if (dictionary.Contains(word.Substring(length, i - length)) &&
Dfs(word, i, visited, dictionary)) {
return true;
}
}
return false;
}
public IList<string> FindAllConcatenatedWordsInADict(string[] words) {
HashSet<string> dictionary = new HashSet<string>(words);
List<string> answer = new List<string>();
foreach (string word in words) {
bool[] visited = new bool[word.Length];
if (Dfs(word, 0, visited, dictionary)) {
answer.Add(word);
}
}
return answer;
}
}C++ 解法
自動ドラフト、提出前に確認#include <bits/stdc++.h>
using namespace std;
// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
class Solution {
public:
private bool Dfs(string word, int length, bool[] visited, HashSet<string> dictionary) {
if (length == word.size()) {
return true;
}
if (visited[length]) {
return false;
}
visited[length] = true;
for (int i = word.size() - (length == 0 ? 1 : 0); i > length; --i) {
if (dictionary.Contains(word.Substring(length, i - length)) &&
Dfs(word, i, visited, dictionary)) {
return true;
}
}
return false;
}
public vector<string> FindAllConcatenatedWordsInADict(vector<string> words) {
HashSet<string> dictionary = new HashSet<string>(words);
List<string> answer = new List<string>();
foreach (string word in words) {
bool[] visited = new bool[word.size()];
if (Dfs(word, 0, visited, dictionary)) {
answer.push_back(word);
}
}
return answer;
}
}Java 解法
照合済み/オリジナルimport java.util.*;
public class Solution {
private boolean dfs(String word, int length, boolean[] visited, Set<String> dictionary) {
if (length == word.length()) {
return true;
}
if (visited[length]) {
return false;
}
visited[length] = true;
for (int i = word.length() - (length == 0 ? 1 : 0); i > length; --i) {
if (dictionary.contains(word.substring(length, i)) && dfs(word, i, visited, dictionary)) {
return true;
}
}
return false;
}
public List<String> findAllConcatenatedWordsInADict(String[] words) {
Set<String> dictionary = new HashSet<>(Arrays.asList(words));
List<String> answer = new ArrayList<>();
for (String word : words) {
boolean[] visited = new boolean[word.length()];
if (dfs(word, 0, visited, dictionary)) {
answer.add(word);
}
}
return answer;
}
}JavaScript 解法
照合済み/オリジナルclass Solution {
dfs(word, length, visited, dictionary) {
if (length === word.length) {
return true;
}
if (visited[length]) {
return false;
}
visited[length] = true;
for (let i = word.length - (length === 0 ? 1 : 0); i > length; i--) {
if (dictionary.has(word.slice(length, i)) && this.dfs(word, i, visited, dictionary)) {
return true;
}
}
return false;
}
findAllConcatenatedWordsInADict(words) {
const dictionary = new Set(words);
const answer = [];
for (const word of words) {
const visited = Array(word.length).fill(false);
if (this.dfs(word, 0, visited, dictionary)) {
answer.push(word);
}
}
return answer;
}
}
// Example usage
const solution = new Solution();
const words = ["cat", "cats", "catsdogcats", "dog", "dogcatsdog", "hippopotamuses", "rat", "ratcatdogcat"];
console.log(solution.findAllConcatenatedWordsInADict(words));Python 解法
照合済み/オリジナルclass Solution:
def dfs(self, word, length, visited, dictionary):
if length == len(word):
return True
if visited[length]:
return False
visited[length] = True
for i in range(len(word) - (1 if length == 0 else 0), length, -1):
if word[length:i] in dictionary and self.dfs(word, i, visited, dictionary):
return True
return False
def findAllConcatenatedWordsInADict(self, words):
dictionary = set(words)
answer = []
for word in words:
visited = [False] * len(word)
if self.dfs(word, 0, visited, dictionary):
answer.append(word)
return answerGo 解法
照合済み/オリジナルpackage main
import (
"strings"
)
type Solution struct{}
func (s Solution) dfs(word string, length int, visited []bool, dictionary map[string]bool) bool {
if length == len(word) {
return true
}
if visited[length] {
return false
}
visited[length] = true
for i := len(word) - 1; i > length; i-- {
if dictionary[word[length:i]] && s.dfs(word, i, visited, dictionary) {
return true
}
}
return false
}
func (s Solution) findAllConcatenatedWordsInADict(words []string) []string {
dictionary := make(map[string]bool)
for _, word := range words {
dictionary[word] = true
}
var answer []string
for _, word := range words {
visited := make([]bool, len(word))
if s.dfs(word, 0, visited, dictionary) {
answer = append(answer, word)
}
}
return answer
}
func main() {
solution := Solution{}
words := []string{"cat", "cats", "catsdogcats", "dog", "dogcatsdog", "hippopotamuses", "rat", "ratcatdogcat"}
result := solution.findAllConcatenatedWordsInADict(words)
for _, word := range result {
println(word)
}
}Algorithm
Для каждого слова в списке:
Построить неявный グラフ, в котором узлы представляют индексы символов в слове, а ребра представляют возможность перехода от одного индекса к другому, если substring между ними является словом из списка.
Использовать Depth-first search (DFS) для проверки, можно ли достигнуть узел с индексом word.length от узла с индексом 0 в グラフе.
Если узел word.length достижим от узла 0, добавить слово в ответ.
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