[Leetcode] Word BreakII

Question:

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.

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"].

----------------------------------------------

Solution:

dfs.

但是需要注意的是，在dfs之前，需要判断这个string能不能被这个dictionary分割（Word Break）。

public class Solution {
    public List<String> wordBreak(String s, Set<String> dict) {
        List<String> result=new ArrayList<String>();
        if(!wordBreakPossible(s,dict)) return result;
        dfs(s,dict,result,"",0);
        return result;
    }

    private void dfs(String s, Set<String> dict, List<String> result,String temp, int start) {
        // TODO Auto-generated method stub
        if(start==s.length())
            result.add(temp);
        else{
            if(start!=0)
                temp+=" ";
            for(int i=start;i<s.length();i++){
                String word=s.substring(start, i+1);
                if(dict.contains(word))
                    dfs(s,dict,result,temp+word,i+1);
            }            
        }
    }

    private boolean wordBreakPossible(String s, Set<String> dict) {
        // TODO Auto-generated method stub
        boolean[] state=new boolean[s.length()+1];
        state[0]=true;
        for(int i=1;i<=s.length();i++){
            for(int j=i-1;j>=0;j--){
                if(state[j]&&dict.contains(s.substring(j, i))){
                    state[i]=true;
                    break;
                }
                
            }
        }
        return state[s.length()];
    }
}

 ----------------------------------------------------------------------

20150221:

又忘了在dfs之前去判断这个string是否可以被这个dictionary分割：

导致出现了TLE:

Last executed input:

"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab", ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]

public class Solution {
    public List<String> wordBreak(String s, Set<String> dict) {
        
        List<String> res=new ArrayList<String>();
        if(s==null||s.length()==0||dict==null||dict.size()==0)
            return res;
        if(!wordBreakPossible(s,dict)) return res;
        dfs(res,s,dict,"");
        return res;
    }
    public void dfs(List<String> res,String s,Set<String> dict,String temp){
        if(s.length()==0){
            
            res.add(temp.trim());
            return;
        }
        for(int i=1;i<=s.length();++i){
            String t=s.substring(0,i);
            if(dict.contains(t)){
                dfs(res,s.substring(i),dict,temp+" "+t);
            }else{
                continue;
            }
        }
    }
    private boolean wordBreakPossible(String s, Set<String> dict) {
        // TODO Auto-generated method stub
        boolean[] state=new boolean[s.length()+1];
        state[0]=true;
        for(int i=1;i<=s.length();i++){
            for(int j=i-1;j>=0;j--){
                if(state[j]&&dict.contains(s.substring(j, i))){
                    state[i]=true;
                    break;
                }
                
            }
        }
        return state[s.length()];
    }
}