标签：leetcode   算法   

问题描述：

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"].

基本思想：

此题可以用上一篇《word break》的思路来解决分词问题。然后在根据分词表格生成可能的分词路径。

本题在生成可能的分词序列时，是从tail到head扫描的。之前我的方法是从head到tail扫描的，但是TLE 错误，有个case过不去，如下：

"aaaaaaaaaaaaaaaaaaaaaaaaaaaa...ab"         ["a","aa","aaa","b"]

但是反过来扫描就ac了。说明test case 中没有形如这样的case：

"baaaaaaaaaaaaaaaaaaaaaaaaaa..aaa"         ["a","aa","aaa","b"] 

如果有更好的方法可以通过两种case，请不吝指教，谢谢～

代码：

 public void getSegStr(int end,String cur,Set<String> dict,String s,List<String> result,int [][] record)  //java
    {
        String item;
        for(int i=end-1; i>=0 ; i--)
        {
            item = cur;
            if(record[i][end]==1 && dict.contains(s.substring(i,end)))
            {
                item = s.substring(i,end)+" "+item;
                if(i==0)
                    result.add(item.trim());
                else
                    getSegStr(i,item,dict,s,result,record);
                    
            }
        }
    }
    public List<String> wordBreak(String s, Set<String> dict) {
        List<String> result = new ArrayList<String>();
        if(s.isEmpty())
    	    return result;
    	if(dict.contains(s))
    	{
    	    result.add(s);
    	    return result;
    	}
    	
    	int len = s.length();
    	int [][] record = new int[len+1][len+1];
    	for(int i=0; i<=len; i++)
    	    record[i][i]=1;
    	
    	for(int step=1; step<=len; step++)
    	{
        	for(int j=step; j<=len; j++)
        	{
        	    int i=j-step;
        	    if(dict.contains(s.substring(i,j)))
        	     {
        	         record[i][j]=1;
        	         continue;
        	     }
        	     
        	     for(int k=i+1; k<j; k++)
        	     {
        	         if(record[i][k]==1 && record[k][j]==1)
        	         {
        	            record[i][j] = 1;   
        	            break;
        	         }
        	     }
        	}
    	}
    
    	//drawback
        getSegStr(s.length(),"",dict,s,result,record);
    	return result;
    }

[leetcode]

标签：leetcode   算法   

原文地址：http://blog.csdn.net/chenlei0630/article/details/42502329