[leetcode]Word Break

问题描述：

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"

.

基本思想：

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。

代码：

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;
}