LeetCode | 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 {
    public boolean wordBreak(String s, Set<String> wordDict) {
        
        if(s.length() == 0) return true;
        if(s.length() == 1) return wordDict.contains(s) ? true : false;
        
        //状态变量：dp[i]代表以第个字符结尾的string，能否用字典词合成
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;      //代表空字串总能合成，否则在下面循环中就得考虑s自身在字典中的情况
        dp[1] = wordDict.contains(s.substring(0,1)) ? true : false;
        
        //状态转移方程：即用dp[0-(i-1)]来确定dp[i]
        //对以第i个字符结尾的字符串来说，若他的前面j个字符的子串能用字典词合成
        //且(j~i)的子串又是字典词的话，那么其就能由字典词合成
        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()];
    }
}