leetcode(word-break)

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”.

解题思路：使用动态规划算法。首先找到前面能够直接作为单词的子串并将它赋值为ｔｒｕｅ。

public boolean wordBreak(String s, Set<String> dict) {      boolean word[]=new boolean[s.length()];//保存前面ｉ长度的字符串能否划分的值        int i;        for(i=0;i<s.length();i++)        if(dict.contains(s.substring(0, i+1)))word[i]=true;        else word[i]=false;        for(i=0;i<s.length();i++){            for(int j=0;j<=i;j++){            if(word[j]&&dict.contains(s.substring(j+1,i+1)))            {//如果前ｊ个字符能够进行划分而且ｊ到ｉ的子串是一个单词则说明前ｊ个字符能够划分                    word[i]=true;                    break;                                  }            }        }        return word[s.length()-1];       }