标签：length duplicate star not substr contains ict please private
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. You may assume the dictionary does not contain duplicate words. 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"]. UPDATE (2017/1/4): The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.
通过hashMap进行记忆化搜索, 不过是从单词表中走, 而非s的长度上遍历
public List<String> wordBreak(String s, List<String> wordDict) {
List<String> res = new LinkedList<String>();
if (wordDict.size() == 0 || wordDict == null) {
return res;
}
return helper(new HashMap<String, LinkedList<String>>(), s, wordDict);
}
private LinkedList helper( HashMap<String, LinkedList<String>> map, String s, List<String> wordDict) {
if (map.containsKey(s)) {
return map.get(s);
}
LinkedList<String> list = new LinkedList<>();
if (s.length() == 0) {
list.add("");
return list;
}
for (String word : wordDict) {
if (s.startsWith(word)) {
String sub = s.substring(word.length());
LinkedList<String> subList = helper(map, sub, wordDict);
for (String item : subList) {
list.add(word + (item.isEmpty() ? "" : " ") + item);
}
}
}
map.put(s, list);
return list;
}
标签：length duplicate star not substr contains ict please private
原文地址：http://www.cnblogs.com/apanda009/p/7231190.html
