LeetCode 140. Word Break II（单词切分）

原题网址：https://leetcode.com/problems/word-break-ii/

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given

s =

"catsanddog"

,

dict =

["cat", "cats", "and", "sand", "dog"]

.

A solution is

["cats and dog", "cat sand dog"]

.
方法一：深度优先。

public class Solution {
private ArrayList<String>[] scanned;
private List<String> results = new ArrayList<>();
private void dfs(String s, int from, List<String> words, Set<String> wordDict) {
if (from == s.length()) {
String result = "";
for(int i=0; i<words.size(); i++) {
if (i>0) result += " ";
result += words.get(i);
}
results.add(result);
return;
}
for(String word: scanned[from]) {
words.add(word);
dfs(s, from + word.length(), words, wordDict);
words.remove(words.size()-1);
}
}
public List<String> wordBreak(String s, Set<String> wordDict) {
scanned = new ArrayList[s.length()];
scanned[0] = new ArrayList<>();
boolean reachable = false;
for(int i=0; i<s.length(); i++) {
if (scanned[i] == null) continue;
for(String word: wordDict) {
if (i + word.length() <= s.length() && word.equals(s.substring(i, i+word.length()))) {
scanned[i].add(word);
if (i+word.length() == s.length()) reachable = true;
if (i+word.length() < s.length() && scanned[i+word.length()] == null) {
scanned[i+word.length()] = new ArrayList<>();
}
}
}
}
if (!reachable) return results;
dfs(s, 0, new ArrayList<>(), wordDict);
return results;
}
}

方法二：Trie+深度优先搜索。

public class Solution {
private TrieNode root;
private void find(char[] sa, int from, int[] split, int splits, List<String> words) {
if (from == sa.length) {
char[] word = new char[sa.length+splits-1];
int wordPos = 0, splitPos = 0;
for(int i=0; i<sa.length; i++) {
if (i==split[splitPos]) { word[wordPos++] = ' '; splitPos++; }
word[wordPos++] = sa[i];
}
words.add(new String(word));
return;
}
TrieNode current = root;
for(int i=from;i<sa.length;i++) {
current = current.nexts[sa[i]-'a'];
if (current == null) break;
if (current.isWord) {
split[splits] = i+1;
find(sa, i+1, split, splits+1, words);
}
}
}
public List<String> wordBreak(String s, Set<String> wordDict) {
List<String> results = new ArrayList<>();
if (s==null || wordDict==null) return results;
root = new TrieNode();
for(String word: wordDict) {
char[] wa = word.toCharArray();
TrieNode current = root;
for(int i=0; i<wa.length; i++) current = current.add(wa[i]);
current.isWord = true;
}
char[] sa = s.toCharArray();
boolean[] reachable = new boolean[sa.length+1];
reachable[0] = true;
for(int i=0; i<sa.length; i++) {
if (!reachable[i]) continue;
TrieNode current = root;
for(int j=i;j<sa.length && current != null;j++) {
current = current.nexts[sa[j]-'a'];
if (current != null && current.isWord) reachable[j+1] = true;
}
}
if (!reachable[sa.length]) return results;
find(sa, 0, new int[sa.length], 0, results);
return results;
}
}
class TrieNode {
boolean isWord;
TrieNode[] nexts = new TrieNode[26];
TrieNode add(char ch) {
int i = ch - 'a';
if (nexts[i] != null) return nexts[i];
nexts[i] = new TrieNode();
return nexts[i];
}
}