leetcode--Word Break II

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Retun all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

 

Have you been asked this question in an interview? 

 

A  clever way:

public class Solution {
	public List<String> wordBreak(String s, Set<String> dict) {
		List<String> result = new ArrayList<String>();
		List<List<Integer> > split = new ArrayList<List<Integer> >();
		boolean[] cuts = new boolean[s.length() + 1];
		if(s.length() > 0 && dict.size() > 0) {
			split.add(0, new ArrayList<Integer>());
			cuts[0] = true;
			for(int i = 1; i < s.length() + 1; ++i) {
				cuts[i] = false;
				List<Integer> currentSplit = new ArrayList<Integer>();
				for(int j = 0; j < i; ++j) {
					if(!cuts[j]) continue;
					else {
						if(dict.contains(s.substring(j, i))){
							if(!cuts[i])
								cuts[i] = true;
							currentSplit.add(j);
						}
					}
				}
				split.add(currentSplit);
			}
			//build splits
			if(!split.get(s.length()).isEmpty())
				buildSplit(s, split,s.length(), result);
		}
		return result;
	}
		
	//dfs method
	private void buildSplit(String s, List<List<Integer> > split, int i, List<String> result){ 
		List<Integer> insertBlank = split.get(i);
		for(int j : insertBlank){
			StringBuffer aCopy = new StringBuffer(s);
			if(j != 0){
				aCopy.insert(j,  " ");
				buildSplit(aCopy.toString(), split, j, result);
			}
			else
				result.add(s);
		}
	}
}

　

In this post, I used a very stupid method to solve this problem:

1. Using the dynamic programming method to find all possible position to cut the string(The same as Word Break)

This step will return a boolean matrix, if(i,j) entry is true then the substring s.substring(i,j) can be splitted.

2. Using the boolean matrix obtained in step1, we do DFS to obtain all strings.

public class Solution {
    public ArrayList<String> wordBreak(String s, Set<String> dict) {
        ArrayList<String> result = new ArrayList<String>();
		if(s == null || dict.size() == 0)
			return result;
		//find the position to split the string
		int len = s.length();
		boolean[][] splits = new boolean[len][len + 1];
		for(int i = len - 1; i >= 0; --i){
			splits[i][i] = false;
			for(int j = i + 1; j < len + 1; ++j){
				if(dict.contains(s.substring(i, j)))
					splits[i][j] = true;
				else{
					for(int k = i; k < j; ++k){
						splits[i][j] = splits[i][k] && splits[k][j];
						if(splits[i][j])
							break;
					}
				}
			}
		}
		//find the possible splitting of the string
		//I use dfs method to solve this.
		if(splits[0][len]){
			ArrayList<ArrayList<Integer>> resultInPosition = new ArrayList<ArrayList<Integer>>();
			HashMap<Integer, ArrayList<ArrayList<Integer>> > finished = new HashMap<Integer,ArrayList<ArrayList<Integer>> >();
		       	for(int i = len - 1; i >= 0; --i){
				ArrayList<ArrayList<Integer>> startAti = new ArrayList<ArrayList<Integer>>();
				if(splits[i][len]){
					if(dict.contains(s.substring(i, len))){
						ArrayList<Integer> aList = new ArrayList<Integer>();
						aList.add(len);
						startAti.add(aList);
					}
					for(int j = i + 1; j < len; ++j){
						if(dict.contains(s.substring(i,j)) && splits[j][len]){
							ArrayList<ArrayList<Integer> > fin = finished.get(j);
							for(ArrayList<Integer> ls : fin){
								ArrayList<Integer> temp = new ArrayList<Integer>();
								temp.add(j);
								temp.addAll(ls);
								startAti.add(temp);
							}			
						}		
					}
				}
				finished.put(i, startAti);
			}
			resultInPosition = finished.get(0);
			for(ArrayList<Integer> ls : resultInPosition){
				int lssize = ls.size();
			       	int first = ls.get(0);	
				String resultString = s.substring(0, first);
				for(int i = 1; i < lssize; ++i){
					resultString += (" "+ s.substring(first, ls.get(i)));
					first = ls.get(i);
				}
				result.add(resultString);			
			}
					
		}
		return result;
    }
}

I tested this code on leetcode online judge, it was accepted and it used 450ms for 24 test cases. I think it is ok although it may not be the best.