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 {
public boolean wordBreak(String s, Set<String> wordDict) {
if(s.length() == 0) return true;
if(s.length() == 1) return wordDict.contains(s) ? true : false;
//狀態變量:dp[i]代表以第個字符結尾的string,能否用字典詞合成
boolean[] dp = new boolean[s.length() + 1];
dp[0] = true; //代表空字串總能合成,否則在下面循環中就得考慮s自身在字典中的情況
dp[1] = wordDict.contains(s.substring(0,1)) ? true : false;
//狀態轉移方程:即用dp[0-(i-1)]來確定dp[i]
//對以第i個字符結尾的字符串來說,若他的前面j個字符的子串能用字典詞合成
//且(j~i)的子串又是字典詞的話,那麼其就能由字典詞合成
for(int i=2; i<dp.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()];
}
}