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".
查看字符串是否由字典中的单词组成。
1、暴力递归，果断超时。
public class Solution {
public boolean wordBreak(String s, Set<String> wordDict) {
return helper(s,0,wordDict);
}
public boolean helper(String s,int start,Set<String> wordDict ){
if( start == s.length() )
return true;
for( int i = s.length();i > start ;i--){
if( wordDict.contains( s.substring(start,i) ) ){
if( helper(s,i,wordDict) )
return true;
}
}
return false;
}
}
2、dp，基本达到最快。
int len = s.length(),maxLen = 0;
boolean[] dp = new boolean[len];
for( String str : wordDict ){
maxLen = Math.max(maxLen,str.length());
} for( int i = 0 ;i<len;i++){ for( int j = i;j>=0 && i-j<maxLen;j-- ){
if( ( j == 0 || dp[j-1] == true ) && wordDict.contains(s.substring(j,i+1)) ){
dp[i] = true;
break;
}
}
} return dp[len-1];