Word Break I,II

Given a string s and a dictionary of words dict, determine ifs can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

 

public class Solution {
    // http://needjobasap.blogspot.com/2014/11/word-break-leetcode.html
    //http://fisherlei.blogspot.com/2013/11/leetcode-word-break-solution.html
    public boolean wordBreak(String s, Set<String> dict) {
        
        boolean[] dp = new boolean[s.length()+1];
        dp[0] = true;
        
      //  for(int i=0;i<s.length();i++){
      for(int i=1;i<=s.length();i++){
            for(int j=0; j< i; j++){
                if(dp[j] && dict.contains(s.substring(j,i))) {
                    dp[i]=true;
                    break;
                }
            }
        }
        return dp[s.length()];
        
    }
}

 

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"].

就是在1 的基础上加了个dfs的应用

public class Solution {
    public ArrayList<String>  wordBreak(String s, Set<String> wordDict) {
        ArrayList<String> res = new ArrayList<String>();
        if(s==null||s.length()==0) return res;
        int len = s.length();
        boolean [] dp = new boolean[len+1];
        dp[0] = true;
        for(int i=1;i<=len;i++){
            for(int j=0;j<i;j++){
                String tmp = s.substring(j,i);
                if(dp[j] && wordDict.contains(tmp)){
                    dp[i] = true;
                    break;
                }
            }
        }
        if(dp[len]==false) return res;
        StringBuilder item  = new StringBuilder();
        dfs(s, 0, item, res, wordDict);
        return res;
    }
    private void dfs(String s, int start, StringBuilder item, ArrayList<String> res, Set<String> dict){
        int len = s.length();
        if(start>=len){
            res.add(new String(item));
            return;
        }
        for(int i=start+1;i<=len;i++){
            String sub = s.substring(start,i);
            if(dict.contains(sub)){
                int ol = item.length();
                if(ol!=0){
                    item.append(" ");
                }
                item.append(sub);
                dfs(s,i,item,res,dict);
                item.delete(ol,item.length());
            }
        }
    }
}