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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".
Dynamic programming。用一个table记录下0~i-1是否符合要求。注意一开始的table[0]=true，以及table长度为输入字符串长度加一。Loop 0~s.length()-1,只能从table[i]=true的地方开始，因为那之前都是匹配的才能接着匹配。
public class Solution { public boolean wordBreak(String s, Set<String> dict) { int strlen=s.length(); boolean [] table=new boolean[strlen+1]; table[0]=true;//begin is always true, so we can start the loop for(int i=0;i<strlen;i++){ if(!table[i])// 0~i-1 are not words continue; for(String word: dict){ int len=word.length(); int end=i+len-1; if(end>strlen-1) continue; if(table[end+1])//already match at this position continue; if(s.substring(i,end+1).equals(word)){ table[end+1]=true; } } } return table[strlen]; } }
Word Break/Word Segment,布布扣,bubuko.com
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原文地址：http://www.cnblogs.com/qingyezi/p/3816404.html
