(待解决)leecode 分词利用词典分词 word break

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不戚戚于贫贱，不汲汲于富贵      ---五柳先生

Given a string s and a dictionary of words dict, determine if s 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".

搜索--》自顶向下的动态规划（备忘录法）》自底向下动态规划

这些思想一直在使用中

1.递归（搜索）

学习递归时候，老师说他的好处 是简单，其实递归的思路是暴力搜索的方法，所以写的人需要考虑的问题很少，但是栈的调用很耗时，同时可能会出现栈溢出。但是他只做自己需要干的事情，，同时他需要干的事情他会重复做。这为后面的自顶向下的优化做了铺垫。

public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        
        if(s=="") return true;
        for(int i=0;i<s.length();i++)
        {
            String s1=s.substring(0,i+1); //枚举所有以0为开头的的字符串
            if(dict.contains(s1)&&wordBreak(s.substring(i+1),dict))
            {
                return true;
                
            }
            
            
        }
        
        return false;
        
        
    }
}

结果是：　　

Last executed input:"acaaaaabbbdbcccdcdaadcdccacbcccabbbbcdaaaaaadb", ["abbcbda","cbdaaa","b","dadaaad","dccbbbc","dccadd","ccbdbc","bbca","bacbcdd","a","bacb","cbc","adc","c","cbdbcad","cdbab","db","abbcdbd","bcb","bbdab","aa","bcadb","bacbcb","ca","dbdabdb","ccd","acbb","bdc","acbccd","d","cccdcda","dcbd","cbccacd","ac","cca","aaddc","dccac","ccdc","bbbbcda","ba","adbcadb","dca","abd","bdbb","ddadbad","badb","ab","aaaaa","acba","abbb"]

 2.超时了，如何优化呢，博主在一本书上看过，是自顶向下动态规划，同学们可以查查，其实就是记住了他重复做的事情。

dp[i][j]表示两者之间是否在词典中。

dp[i][j]=dp[i][k] && dp[k+1][j] (i+1=<k<j-1)

public class Solution {
    
  
    // 
    int dp(int i,int j,String s,Set<String> dict,int d[][]) // get the i to j is exit in the dict
    {
        if(d[i][j]==1) return 1;
        if(d[i][j]==-1) return -1;
        if(dict.contains(s.substring(i,j+1)))
        {
            d[i][j]=1;
            return 1;
            
        }
        else
        {
            for(int k=i;k<j;k++)
            {
                if(dp(i,k,s,dict,d)==1&&dp(k+1,j,s,dict,d)==1)
                {
                    d[i][j]=1;
                    return 1;
                    
                    
                }
                
                
            }
            
            return -1;
            
        }
        
        
    }
    public boolean wordBreak(String s, Set<String> dict) {
        int len=s.length();
        
        int d[][]=new int[len][len];
        int ans=dp(0,len-1,s,dict,d);
        if(ans==1) return true;
         return false;
       
        
        
        
        
    }
}


错了，还是超时，

3.最后的自定向下方法了，填表，那些专业人士都叫打表。

(待解决)leecode 分词利用词典分词 word break,布布扣,bubuko.com

(待解决)leecode 分词利用词典分词 word break

标签：style   blog   java   color   使用   问题   

原文地址：http://www.cnblogs.com/hansongjiang/p/3821212.html