139. Word Break

My BFS Solution 

    public boolean wordBreak(String s, List<String> wordDict) {
        if(wordDict.contains(s))
            return true;
        Queue<String> queue = new LinkedList<>();
        for(String word:wordDict){
            if(s.startsWith(word)){
                queue.add(word);
            }
        }
        while(!queue.isEmpty()){
            int size = queue.size();
            for(int i=0;i<size;i++){
                String tempS = queue.poll();
                String subS = s.substring(tempS.length());
                if(wordDict.contains(subS))
                    return true;
                else
                {
                    for(String word:wordDict){
                        if(subS.startsWith(word)){
                            queue.offer(tempS+word);
                        }
                    }
                }
            }
        }
        return false;
    }

My Backtracking Solution:

   public boolean wordBreak(String s, List<String> wordDict) {
        Set<String> wordSet = new HashSet<>();
        for(String word: wordDict){
            wordSet.add(word);
        }
        return helper(s, wordSet);
    }
    
    private boolean helper(String s, Set<String> set){
        if(s.length()==0)
            return true;
        
        for(String word: set){
            if(s.startsWith(word)){
                if(helper(s.substring(word.length()), set))
                    return true;
            }
        }
        
        return false;
    } 

以上两个Solution虽然是对的，但是对于下面的test case都TLE

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

标签：queue,set,word,String,Break,Word,wordDict,return,139
来源： https://www.cnblogs.com/feiflytech/p/15690934.html