140. Word Break II

yPhantom 2019年10月19日 32次浏览

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:
[]

Solution

单词拆分,要求将给定的字符串按照字典数组中给的单词进行拆分,返回所有可能的结果。这题相比较上一个只需要返回ture或false而言又加大了的难度。

还是得用一个hashmap来作为记忆数组,避免tle.

class Solution {
    
    private Map<String, List<String>> map = new HashMap<>();

    public List<String> wordBreak(String s, List<String> wordDict) {
        if (map.containsKey(s)) {
            return map.get(s);
        }
        List<String> res = new LinkedList<>();
        if (s.length() == 0) {
            res.add("");
            return res;
        }
        for (String word: wordDict) {
            if (s.startsWith(word)) {
                List<String> subList = wordBreak(s.substring(word.length()), wordDict);
                for (String sub: subList) {
                    res.add(word + (sub.isEmpty() ? "" : " ") + sub);
                }
            }
        }
        map.put(s, res);
        return res;
    }
}