[leedcode 139] 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"
.
public class Solution { //DP思想,状态量dp[i],表示字符串s的前i个字符组成的子串是否能够进行单词拆分。 //dp[0]=false; //如果一个字符串 s[0,i) 能被拆分,则一定能找到一个j使得:s[0,j) 能被拆分 且 s[j,i) 在字典中 public boolean wordBreak(String s, Set<String> wordDict) { boolean dp[]=new boolean[s.length()+1]; dp[0]=true; for(int i=1;i<=s.length();i++){ for(int j=0;j<i;j++){ if(dp[j]&&wordDict.contains(s.substring(j,i))){ dp[i]=true; break; } } } return dp[s.length()]; } }