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:
[]

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。

public class Solution {
    HashMap<String,List<String>> map = new HashMap<String,List<String>>();
    public List<String> wordBreak(String s, List<String> wordDict) {
        if(map.containsKey(s)) {
            return map.get(s);
        }
        List<String> res = new ArrayList<String>();
        if(s == null || s.length() == 0) {
            return res;
        }
        
        if(wordDict.contains(s)) {
            res.add(s);
        }
        for(int i = 1 ; i < s.length() ; i++) {
            String t = s.substring(i);
            if(wordDict.contains(t)) {
                List<String> temp = wordBreak(s.substring(0 , i) , wordDict);
                if(temp.size() != 0) {
                    for(int j = 0 ; j < temp.size() ; j++) {
                        res.add(temp.get(j) + " " + t);
                    }
                }
            }
        }
        map.put(s , res);
        return res;
    }
}

dfs + memorization ≈ dp

用hashmap存已有的string对应的list。然后开始找分割点，分割点前继续dfs，分割点后直接检查是否存在于dict中。

从后往前，如果这个屁股string存在于dict中，就找它前面这个整串的可能的解，比如得到了temp，那我们就需要对每个temp（每种可能）后面加上屁股string，最后更新map和返回当前的解。

 

https://www.youtube.com/watch?v=JqOIRBC0_9c&ab_channel=HuaHua

 从后往前

 

public class Solution {
    public List<String> wordBreak(String s, List<String> wordDict) {
        return DFS(s, wordDict, new HashMap<String, LinkedList<String>>());
    }       

    // DFS function returns an array including all substrings derived from s.
    List<String> DFS(String s, List<String> wordDict, HashMap<String, LinkedList<String>>map) {
        if (map.containsKey(s)) 
            return map.get(s);

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

从前往后