Word Break----leetcode

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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".

题目大意：很浅显，这里不解释了，采用官网上的动态规划的解法，注意可能会出现数组越界，区别c++中的substr（pos,len）和java中的substring（beginIndex，endIndex），参数：beginIndex:开始处的索引（包括），endIndex :结束处的索引（不包括）。代码如下：

public class Solution {    public boolean wordBreak(String s, Set<String> dict) {        if(dict.isEmpty()) {            return false;        }        if(s.length()==0)             return true;        int length;        length=s.length();        boolean[] wordB=new boolean[length+1];                wordB[0]=true;        for(int i=1;i<=length;i++)            for(int j=0;j<i;j++)            {                if(wordB[j]&&dict.contains(s.substring(j,i)))                {                    wordB[i]=true;                    break;                }            }        return wordB[length];    }}

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