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".
动态规划思想: 保存每个子串S(i,j)是否可分的信息。从小到大构建可分性表格。
public boolean wordBreak(String s, Set<String> dict) { //java
if(s.isEmpty())
return true;
if(dict.contains(s))
return true;
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;
}
}
}
}
if(record[0][len] == 1) return true;
else return false;
}原文:http://blog.csdn.net/chenlei0630/article/details/42468687