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.

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"].

 

public class Solution {
    public List<String> wordBreak(String s, Set<String> wordDict) {
        ArrayList<String> res = new ArrayList<>();
        if (s == null || s.length() == 0) {
        	return res;
        }
        if (wordBreakcheck(s, wordDict)) {
        	solve(s, wordDict, 0, "", res);
        }
        return res;
    }

	private void solve(String s, Set<String> wordDict, int start, String item,
			ArrayList<String> res) {
		if (start >= s.length()) {
			res.add(item);
			return;
		}
		StringBuffer stringBuffer = new StringBuffer();
		for (int i = start; i < s.length(); i++) {
			stringBuffer.append(s.charAt(i));
			if (wordDict.contains(stringBuffer.toString())) {
				String tmp = new String();
				if (item.length() > 0) {
					tmp = item + " " + stringBuffer.toString();
				} else {
					tmp = stringBuffer.toString();
				}
				solve(s, wordDict, i+1, tmp, res);
			}
		}
	}

	private boolean wordBreakcheck(String s, Set<String> wordDict) {
		if (s == null || s.length() == 0) {
			return true;
		}
		boolean[] dp = new boolean[s.length()+1];
		dp[0] = true;
		for (int i = 0; i < s.length(); i++) {
			StringBuffer stringBuffer = new StringBuffer(s.substring(0, i+1));
			for (int j = 0; j <= i; j++) {
				if (dp[j] && wordDict.contains(stringBuffer.toString())) {
					dp[i+1] = true;
					break;
				}
				stringBuffer.deleteCharAt(0);
			}
		}
		return dp[s.length()];
	}
}