Leetcode 139. Word Break I&&II

最新推荐文章于 2025-04-13 18:39:57 发布

ld851

最新推荐文章于 2025-04-13 18:39:57 发布

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分类专栏：编程题文章标签： LeetCode 字符串

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本文探讨了LeetCode上的Word Break I 和 Word Break II两个问题的解决方案，包括递归和动态规划的方法，并提供了完整的Java代码实现。

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题目来源：leetcode

Word Break I

Given a non-empty string s and a dictionary wordDict containing a list of non-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 of non-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()];
     }
 }