LeetCode 140. Word Break II
Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.
Note:
The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words.
Example 1:
Input:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
"cats and dog",
"cat sand dog"
]
Example 2:
Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
"pine apple pen apple",
"pineapple pen apple",
"pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.
Example 3:
Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]
不解释,代码如下:
public List<String> wordBreak(String s, List<String> wordDict) {
HashSet<String> set = new HashSet<>();
for (String ss : wordDict) {
set.add(ss);
}
boolean[] breakable = new boolean[s.length() + 1];
breakable[0] = true;
for (int i = 0; i < s.length(); i++) {
for (int p = 0; p <=i; p++) {
String curS = s.substring(p, i+1);
if (set.contains(curS) && breakable[p]) {
breakable[i+1] = true;
}
}
}
if (!breakable[s.length()]) {
return new ArrayList<>();
}
List<List<String>> allResult = new ArrayList<>();
for (int i = 0; i < s.length(); i++) {
List<String> res = new ArrayList<>();
for (int p = 0; p <= i; p++) {
String curs = s.substring(p, i + 1);
if (set.contains(curs)) {
if (p == 0) {
res.add(curs);
} else if (allResult.get(p-1).size() != 0) {
for (String ss : allResult.get(p-1)) {
ss = ss + " " + s.substring(p, i + 1);
res.add(ss);
}
} else {
continue;
}
}
}
allResult.add(res);
}
return allResult.get(allResult.size() - 1);
}