题目
Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
Note:
- The same word in the dictionary may be reused multiple times in the segmentation.
- You may assume the dictionary does not contain duplicate words.
Example 1:
Input: s = "leetcode", wordDict = ["leet", "code"] Output: true Explanation: Return true because"leetcode"
can be segmented as"leet code"
.
Example 2:
Input: s = "applepenapple", wordDict = ["apple", "pen"] Output: true Explanation: Return true because"
applepenapple"
can be segmented as"
apple pen apple"
. Note that you are allowed to reuse a dictionary word.
Example 3:
Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"] Output: false
十分钟尝试
尝试用dp思路去解决,但是不知道如何写转化方程。
dp[i]如果满足条件,需要dp[i-1]满足条件,如何定义呢?看下面的关键点,如果f[j]可以由字典单词组成,并且s[j,i](左闭右开)在字典中国,那么f[i]就是可以组成,他的值就是true。
f[j] is to check if s[0:j] (0 inclusive, j exclusive) can be composed of words in wordDict, if both f[j] is true and s[j:i] (j inclusive, i exclusive) is in wordDict, then we can say f[i] is true. (Note that f's length is 1 more than s's length, which can cause some confusion
class Solution {
public boolean wordBreak(String s, List<String> wordDict) {
boolean[] dp=new boolean[s.length()+1];
Set<String> set=new HashSet(wordDict);
dp[0]=true;
for(int i=1;i<=s.length();i++){
for(int j=0;j<i;j++){
if(dp[j]&&set.contains(s.substring(j,i))){
dp[i]=true;
break;
}
}
}
return dp[s.length()];
}
}