[leetcode]break word II

问题描述：

Given a string s and a dictionary of wordsdict, 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;
}