140. Word Break II

Posted 2020-12-01 wentiliangkaihua

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:
[]

class Solution {
    public List<String> wordBreak(String s, List<String> wordDict) {
        List<String> res = new ArrayList();
        int l = s.length();
        boolean[] dp = new boolean[l + 1];
        dp[0] = true;
        for(int i = 1; i <= l; i++){
            for(int j = 0; j < i; j++){
                if(dp[j] && wordDict.contains(s.substring(j, i))){
                    dp[i] = true;
                }
            }
        }
        if(dp[s.length()] == false)   return res;
        help(s, wordDict, res, 0, "");
        return res;
    }
    public void help(String s, List<String> wordDict, List<String> res, int start, String tmp){
        if(start == s.length()){
            res.add(tmp);
            return;
        }
        if(tmp.length() != 0) tmp += " ";
        for(int i = start; i < s.length(); i++){
            String word = s.substring(start, i+1);
            if(!wordDict.contains(word)) continue;
            help(s, wordDict, res, i + 1, tmp + word);
        }
    }
}

需要先判断是否能进行word break再通过dfs求path。