Leetcode 139. Word Break I&&II

题目来源：leetcode

Word Break I 

Word Break I 

Given a non-empty string s and a dictionary wordDict containing a list ofnon-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.

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

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

思路一：递归（大数据超时）

class Solution {    public boolean wordBreak(String s, List<String> wordDict) {                boolean res=false;        if(s.length()==0)            return false;        char c=s.charAt(0);        for(String seq:wordDict) {            if (seq.length() <= s.length()) {                if (c == seq.charAt(0)) {                    int i = 0;                    for (; i < seq.length(); i++) {                        if ( s.charAt(i) != seq.charAt(i))                            break;                    }                     if (i==s.length())                         return true;                    else if (i == seq.length()) {                        res = wordBreak(s.substring(seq.length(), s.length()), wordDict);                        if(res==true)                            return res;                    }
                }            }        }        return res;
    }}

思路二：动态规划；dp记录每加入一个字符的是否可以分割  （怎么想不到呢？？）

           

class Solution {    public boolean wordBreak(String s, List<String> wordDict) {        boolean dp[]=new boolean[s.length()+1];        dp[0]=true;        for(int i=1;i<=s.length();i++)            for(int j=0;j<i;j++){                if(dp[j]&&wordDict.contains(s.substring(j,i)))                {dp[i]=true;                break;                }            }        return dp[s.length()];            }}

Word Break I I

Given a non-empty string s and a dictionary wordDict containing a list ofnon-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. You may assume the dictionary does not contain duplicate words.

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"].

思路：列举出来递归   没A过，不想改了

/** * Created by a819 on 2017/8/19. */import java.util.*;public class wordBreakII {    public List<String> wordBreak(String s, List<String> wordDict) {        ArrayList<String> res = new ArrayList<>();        StringBuilder sb=new StringBuilder();        //boolean res=false;        if (s.length() == 0)            return res;        res=wordBreakI(s,wordDict,sb,res);      /* for(int i=res.size()-1;i>=0;i--){           res.add(res.get(i));       }*/        return res;   }    public ArrayList<String> wordBreakI(String s, List<String> wordDict, StringBuilder sb,ArrayList<String> ls) {        char c = s.charAt(0);        for (String seq : wordDict) {            if (seq.length() <= s.length()) {                if (c == seq.charAt(0)) {                    int i = 0;                    for (; i < seq.length(); i++) {                        if (s.charAt(i) != seq.charAt(i))                            break;                    }                    if (i == s.length()) {                        sb.append(" "+seq);                        sb.delete(0,1);                        ls.add(sb.toString());                        //sb=new StringBuilder();                        return ls;                    }                    else if (i == seq.length()) {                        sb.append(" "+seq);                        ls= wordBreakI(s.substring(seq.length(), s.length()), wordDict,sb,ls);                        sb=new StringBuilder();//每生成一个片段要及时清空，否则会加到下一个片段                    }                }            }        }        return ls;    }    public static void main(String[] args) {        String s="aaaaaaa";        String []a=new String[]{"aaaa","aa","a"};        wordBreakII w=new wordBreakII();        System.out.println(w.wordBreak(s,Arrays.asList(a)).size());    }    }

正确：而且需要剪纸

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()];     } }